Algebraic Formulae & Identities — Complete Guide with Tricks, Derivations & Practice | Jnaanangkur
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Algebraic Formulae & Identities, Explained the Simple Way

Every identity from (a+b)² to a³+b³+c³−3abc — with derivations, solved examples, memory tricks, common mistakes, and exam-ready practice. Built for NCERT, CBSE and State Board (SEBA/Assam Board) students of Classes 8–10.

👋 Welcome, Dear Students ! If algebraic identities feel like a jumble of letters and signs right now, take a breath — by the end of this guide, you'll be able to look at (a+b)² or a³−b³ and instantly know what it expands to, why it works, and how to use it to solve problems faster than your classmates. Let's build that confidence together, one formula at a time.
✅ 10 Core Identities ✅ 30+ Solved Examples ✅ Practice + Answer Keys ✅ NCERT • CBSE • SEBA Aligned
Getting Started

What Exactly Is an Algebraic Identity?

Let's clear this up before anything else, because most confusion in algebra starts right here.

An algebraic equation is true only for certain values of the variable. For example, x + 2 = 5 is true only when x = 3.

An algebraic identity, on the other hand, is true for every possible value of the variables involved. For example, (a+b)² = a² + 2ab + b² is true whether a = 2, a = −7, a = 0.5, or anything else — you can plug in any numbers and both sides will always match.

Think of an identity as a verified shortcut — one that mathematicians have already proven works every single time, so you don't have to multiply everything out longhand again and again.

That's the entire point of this guide: once you trust that an identity always works, you can use it to expand expressions in seconds, factorise tricky polynomials, and even do lightning-fast mental maths (like squaring 105 without a calculator).

See It, Don't Just Memorise It

The Geometric Proof of (a+b)²

Drag the sliders below and watch why (a+b)² is never just a² + b². The square's area is built from four smaller pieces — and that's the whole secret of the identity.

ab
ab
(6 + 4)² = 6² + 2(6)(4) + 4² = 100

The big square of side (a+b) is made of one a×a square, one b×b square, and two a×b rectangles. Add their areas — that's exactly the identity!

Print This. Stick It On Your Wall.

📊 Master Formula Chart — All Identities at a Glance

Your one-stop revision sheet. Bookmark this page and scroll back here the night before your exam.

#IdentityFormulaBest used for
1Square of a sum(a+b)² = a²+2ab+b²Expanding, mental squares
2Square of a difference(a−b)² = a²−2ab+b²Expanding, mental squares
3Difference of squares(a+b)(a−b) = a²−b²Fast multiplication, factorising
4Cube of a sum(a+b)³ = a³+3a²b+3ab²+b³Expanding cubes
5Cube of a difference(a−b)³ = a³−3a²b+3ab²−b³Expanding cubes
6Sum of cubesa³+b³ = (a+b)(a²−ab+b²)Factorising
7Difference of cubesa³−b³ = (a−b)(a²+ab+b²)Factorising
8Square of a trinomial(a+b+c)² = a²+b²+c²+2ab+2bc+2ca3-variable expansion
9Sum of three cubesa³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)Factorising, special case a+b+c=0
10Product of two binomials(x+a)(x+b) = x²+(a+b)x+abQuick multiplication, quadratic forming
The Deep Dive

Every Identity, Fully Explained

For each formula: what it means, how it's derived, worked examples, a memory trick, the mistakes examiners see most often, and a quick check for you to try.

Identity 1 of 10

Square of a Sum

(a + b)² = a² + 2ab + b²

In Plain Language

When you square the sum of two terms, you don't just square each term separately — you also pick up a "cross term" of double their product. This is the single most-used identity in all of school algebra, so make this one automatic.

Derivation

(a+b)² = (a+b) × (a+b)

= a(a+b) + b(a+b)

= a² + ab + ab + b²

= a² + 2ab + b²

Notice the two ab terms are identical, so they combine into 2ab — that's where the "double product" comes from.

Solved Examples

Q1. Expand (3x + 4y)².
Here a = 3x, b = 4y.
a² = 9x², 2ab = 2(3x)(4y) = 24xy, b² = 16y².
Answer: 9x² + 24xy + 16y²
Q2. Find 105² without long multiplication.
Write 105 = 100 + 5. Then 105² = (100+5)² = 100² + 2(100)(5) + 5² = 10000 + 1000 + 25.
Answer: 11025
Q3. If a + b = 10 and ab = 21, find a² + b².
From the identity, a² + b² = (a+b)² − 2ab = 10² − 2(21) = 100 − 42.
Answer: 58

Memory Trick

Say it like a rhythm: "First squared, plus twice the product, plus second squared." Or remember FIRST² + 2(FIRST·SECOND) + SECOND² — the pattern never changes no matter what a and b are.

Common Mistakes

  • (a+b)² = a² + b²Wrong! You must include the middle term 2ab. This is the #1 mistake examiners report.
  • Forgetting to double the product: writing a² + ab + b² instead of a² + 2ab + b².
  • Sign error when a or b is itself negative — always substitute the full term including its sign before squaring.

Quick Check: Expand (2p + 5q)²

a = 2p, b = 5q → 4p² + 2(2p)(5q) + 25q² = 4p² + 20pq + 25q²
Identity 2 of 10

Square of a Difference

(a − b)² = a² − 2ab + b²

In Plain Language

Almost identical to Identity 1, except the cross term is now subtracted. The squared terms a² and b² stay positive (squares are never negative!) — only the middle term changes sign.

Derivation

(a−b)² = (a−b) × (a−b)

= a(a−b) − b(a−b)

= a² − ab − ab + b²

= a² − 2ab + b²

Two negative ab terms combine to give −2ab, while (−b)×(−b) = +b² stays positive.

Solved Examples

Q1. Expand (5p − 3q)².
a = 5p, b = 3q → a² = 25p², 2ab = 30pq, b² = 9q².
Answer: 25p² − 30pq + 9q²
Q2. Find 98² mentally.
Write 98 = 100 − 2. Then 98² = (100−2)² = 10000 − 2(100)(2) + 4 = 10000 − 400 + 4.
Answer: 9604
Q3. If x − y = 6 and xy = 7, find x² + y².
x² + y² = (x−y)² + 2xy = 36 + 14.
Answer: 50

Memory Trick

Same chant as before, just flip one sign: "First squared, minus twice the product, plus second squared." The b² is ALWAYS positive — students often wrongly make it negative because the formula has a minus sign in it.

Common Mistakes

  • Writing a² − 2ab − b² instead of a² − 2ab + b² — the last term is always added, never subtracted.
  • Treating (a−b)² as a² − b² (that's actually a completely different identity — see Identity 3 below!).
  • Sign confusion when a or b is negative, e.g., for (x − (−3))², simplify inside the bracket first to (x+3)² before applying the formula.

Quick Check: Expand (7m − 2n)²

a = 7m, b = 2n → 49m² − 2(7m)(2n) + 4n² = 49m² − 28mn + 4n²
Identity 3 of 10

Difference of Squares

(a + b)(a − b) = a² − b²

In Plain Language

This is the "magic disappearing" identity. When you multiply a sum and a difference of the same two terms, the middle terms cancel out completely, leaving only a difference of two squares. It's the fastest identity for mental multiplication.

Derivation

(a+b)(a−b) = a(a−b) + b(a−b)

= a² − ab + ab − b²

= a² − b²

The −ab and +ab are exact opposites, so they cancel — that's the entire trick.

Solved Examples

Q1. Find 104 × 96 mentally.
Write as (100+4)(100−4) = 100² − 4² = 10000 − 16.
Answer: 9984
Q2. Factorise 49x² − 81y².
49x² = (7x)², 81y² = (9y)² → (7x)² − (9y)².
Answer: (7x + 9y)(7x − 9y)
Q3. Simplify (x + 7)(x − 7).
Directly apply the identity with a = x, b = 7.
Answer: x² − 49

Memory Trick

"Same, same, opposite signs → squares subtract." Whenever you see the exact same two terms, once added and once subtracted, skip the full expansion entirely and jump straight to a² − b². This single identity can save you 30+ seconds per question in MCQs.

Common Mistakes

  • Applying this identity when the terms aren't actually identical, e.g. wrongly simplifying (x+3)(x−5) as x²−15 — this identity ONLY works when both brackets have the same two terms.
  • Writing a² + b² instead of a² − b² — the result is always a subtraction.
  • Forgetting to take the square root correctly while factorising, e.g. mistaking 49 for 7² but writing it as 14 by mistake.

Quick Check: Simplify (9a + 2b)(9a − 2b)

a = 9a, b = 2b → (9a)² − (2b)² = 81a² − 4b²
Identity 4 of 10

Cube of a Sum

(a + b)³ = a³ + 3a²b + 3ab² + b³

In Plain Language

Now we're cubing instead of squaring, so there are four terms instead of three. Notice the pattern in the coefficients: 1, 3, 3, 1 — the powers of a go down (3,2,1,0) while powers of b go up (0,1,2,3) as you move left to right.

Derivation

(a+b)³ = (a+b)² × (a+b)

= (a² + 2ab + b²)(a+b)

= a³ + a²b + 2a²b + 2ab² + ab² + b³

= a³ + 3a²b + 3ab² + b³

There's also a handy shortcut form: (a+b)³ = a³ + b³ + 3ab(a+b) — useful when you already know a+b and ab.

Solved Examples

Q1. Expand (x + 2)³.
a = x, b = 2 → x³ + 3x²(2) + 3x(4) + 8 = x³ + 6x² + 12x + 8.
Answer: x³ + 6x² + 12x + 8
Q2. Find 11³ using the identity.
Write 11 = 10 + 1. (10+1)³ = 1000 + 3(100)(1) + 3(10)(1) + 1 = 1000+300+30+1.
Answer: 1331
Q3. If a+b = 5 and ab = 6, find a³+b³.
Use a³+b³ = (a+b)³ − 3ab(a+b) = 5³ − 3(6)(5) = 125 − 90.
Answer: 35

Memory Trick

Remember the coefficient pattern 1-3-3-1 (it's actually Pascal's Triangle row for power 3!). Pair that with the rule: a's power falls, b's power rises, every term's powers add up to 3.

Common Mistakes

  • The biggest one: writing (a+b)³ = a³ + b³ — this is true ONLY for the special case in Identity 9 (when a+b+c=0), never in general. Always include the two middle terms.
  • Forgetting the coefficient 3 on the middle terms: writing a²b instead of 3a²b.
  • Mixing up which term gets squared — it's 3a²b (a squared) then 3ab² (b squared), not the reverse.

Quick Check: Expand (y + 3)³

a = y, b = 3 → y³ + 3y²(3) + 3y(9) + 27 = y³ + 9y² + 27y + 27
Identity 5 of 10

Cube of a Difference

(a − b)³ = a³ − 3a²b + 3ab² − b³

In Plain Language

The same 1-3-3-1 pattern as before, but now the signs alternate: plus, minus, plus, minus. Watch this one carefully — it's the formula students lose the most marks on due to sign slips.

Derivation

(a−b)³ = (a−b)² × (a−b)

= (a² − 2ab + b²)(a−b)

= a³ − a²b − 2a²b + 2ab² + ab² − b³

= a³ − 3a²b + 3ab² − b³

Shortcut form: (a−b)³ = a³ − b³ − 3ab(a−b).

Solved Examples

Q1. Expand (x − 3)³.
a = x, b = 3 → x³ − 3x²(3) + 3x(9) − 27 = x³ − 9x² + 27x − 27.
Answer: x³ − 9x² + 27x − 27
Q2. Find 9³ using the identity.
Write 9 = 10 − 1. (10−1)³ = 1000 − 3(100)(1) + 3(10)(1) − 1 = 1000−300+30−1.
Answer: 729
Q3. If a−b = 4 and ab = 5, find a³−b³.
Use a³−b³ = (a−b)³ + 3ab(a−b) = 4³ + 3(5)(4) = 64 + 60.
Answer: 124

Memory Trick

Use the phrase "PLUS MINUS PLUS MINUS" like a heartbeat as you write each of the four terms left to right. If you remember Identity 4 correctly, just alternate the signs starting with plus.

Common Mistakes

  • Getting the sign pattern wrong — it is +, −, +, −, never −, +, −, + or all negative.
  • Writing the last term as +b³ instead of −b³ — odd powers of a negative number stay negative.
  • Confusing this with a³ − b³ (Identity 7) — that one has only TWO terms inside one bracket, this one expands into FOUR terms.

Quick Check: Expand (2m − 1)³

a = 2m, b = 1 → 8m³ − 3(4m²)(1) + 3(2m)(1) − 1 = 8m³ − 12m² + 6m − 1
Identity 6 of 10

Sum of Cubes (Factorised Form)

a³ + b³ = (a + b)(a² − ab + b²)

In Plain Language

This goes the opposite direction from Identity 4 — instead of expanding a cube, we're factorising a sum of two cubes back into a product. The second bracket looks similar to a perfect square but has a minus in the middle instead of a plus.

Derivation

Start from the cube identity: (a+b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a+b)

⇒ a³ + b³ = (a+b)³ − 3ab(a+b)

= (a+b)[(a+b)² − 3ab]

= (a+b)[a² + 2ab + b² − 3ab]

= (a+b)(a² − ab + b²)

Solved Examples

Q1. Factorise 8x³ + 27y³.
8x³ = (2x)³, 27y³ = (3y)³ → a = 2x, b = 3y.
Answer: (2x + 3y)(4x² − 6xy + 9y²)
Q2. If x+y = 4 and xy = 3, find x³+y³.
x²+y² = (x+y)²−2xy = 16−6 = 10. So a²−ab+b² = 10−3 = 7.
x³+y³ = (x+y)(7) = 4 × 7.
Answer: 28
Q3. Factorise 125 + a³.
125 = 5³ → a-term is 5, b-term is a.
Answer: (5 + a)(25 − 5a + a²)

Memory Trick

Say: "SAME sign outside, ALWAYS minus in the middle, ALWAYS plus at the end." The first bracket (a+b) keeps the same sign as the original problem, and the second bracket a²−ab+b² ALWAYS has a minus in the middle — no matter what.

Common Mistakes

  • Writing the second bracket as a² + ab + b² instead of a² − ab + b² — this is the most common slip and it's the opposite identity (Identity 7's bracket).
  • Forgetting to express numbers as perfect cubes first — always check if 8, 27, 64, 125, 1000 etc. are cubes before applying the formula.
  • Confusing this with (a+b)³ — remember, this identity factorises, it doesn't expand.

Quick Check: Factorise 1 + 64p³

a = 1, b = 4p → (1 + 4p)(1 − 4p + 16p²)
Identity 7 of 10

Difference of Cubes (Factorised Form)

a³ − b³ = (a − b)(a² + ab + b²)

In Plain Language

The mirror image of Identity 6. This time the first bracket carries the minus sign, but — and this trips up almost everyone — the second bracket flips to a PLUS in the middle.

Derivation

Start from: (a−b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a−b)

⇒ a³ − b³ = (a−b)³ + 3ab(a−b)

= (a−b)[(a−b)² + 3ab]

= (a−b)[a² − 2ab + b² + 3ab]

= (a−b)(a² + ab + b²)

Solved Examples

Q1. Factorise 64x³ − 1.
64x³ = (4x)³, 1 = 1³ → a = 4x, b = 1.
Answer: (4x − 1)(16x² + 4x + 1)
Q2. If x−y = 3 and xy = 4, find x³−y³.
x²+y² = (x−y)²+2xy = 9+8 = 17. So a²+ab+b² = 17+4 = 21.
x³−y³ = (x−y)(21) = 3 × 21.
Answer: 63
Q3. Factorise 1000 − a³.
1000 = 10³ → a-term is 10, b-term is a.
Answer: (10 − a)(100 + 10a + a²)

Memory Trick

Pair the two cube identities side by side: for SUM of cubes, the second bracket is minus in the middle; for DIFFERENCE of cubes, the second bracket is plus in the middle. The signs literally swap between Identity 6 and 7 — learn them together, never alone.

Common Mistakes

  • Writing a² − ab + b² instead of a² + ab + b² in the second bracket — students often just copy Identity 6's bracket by habit.
  • Mixing up a and b when the difference is written "backwards" — e.g. for 1 − 8x³, set a = 1, b = 2x, don't flip them.
  • Stopping after finding (a−b) and forgetting to write the trinomial bracket — both factors are needed for full marks.

Quick Check: Factorise 27p³ − 8

a = 3p, b = 2 → (3p − 2)(9p² + 6p + 4)
Identity 8 of 10

Square of a Trinomial

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca

In Plain Language

Now we have three terms instead of two. You square each one individually, then add the double-product of every possible pair — that's ab, bc, and ca. Miss even one pair and your answer will be wrong.

Derivation

Treat (b+c) as a single term and apply Identity 1: (a+(b+c))² = a² + 2a(b+c) + (b+c)²

= a² + 2ab + 2ac + b² + 2bc + c²

= a² + b² + c² + 2ab + 2bc + 2ca

Solved Examples

Q1. Expand (x+y+2)².
a=x, b=y, c=2 → x²+y²+4 + 2xy + 4y + 4x.
Answer: x²+y²+4+2xy+4x+4y
Q2. If a+b+c = 9 and ab+bc+ca = 26, find a²+b²+c².
a²+b²+c² = (a+b+c)² − 2(ab+bc+ca) = 81 − 52.
Answer: 29
Q3. Expand (2x − y + 3)².
Treat as a=2x, b=−y, c=3 → 4x²+y²+9 + 2(2x)(−y) + 2(−y)(3) + 2(3)(2x).
Answer: 4x²+y²+9−4xy−6y+12x

Memory Trick

Picture a triangle with corners a, b, c. Square each corner, then draw all 3 connecting edges and double each one: 2ab + 2bc + 2ca. Three corners, three edges — nothing left out.

Common Mistakes

  • The most common error: only writing 2ab and forgetting 2bc and 2ca — all three cross-pairs are required.
  • Losing track of signs when b or c is negative — substitute the term with its sign first, then square/multiply.
  • Writing 2(a+b+c) instead of the three separate double-products — they cannot be combined that simply.

Quick Check: Expand (p + 2q − r)²

a=p, b=2q, c=−r → p² + 4q² + r² + 4pq − 4qr − 2pr
Identity 9 of 10

Sum of Three Cubes (Special Identity)

a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)

In Plain Language

This looks intimidating but it's mostly used for one famous special case: if a+b+c = 0, then a³+b³+c³ = 3abc. That single shortcut shows up again and again in HOTS and competitive exam questions.

Derivation (of the special case)

From the identity: a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca)

If a+b+c = 0, the right-hand side becomes 0 (since anything multiplied by 0 is 0):

⇒ a³+b³+c³ − 3abc = 0

⇒ a³+b³+c³ = 3abc

Solved Examples

Q1. Verify the special case using a=1, b=2, c=−3 (note 1+2−3=0).
a³+b³+c³ = 1+8−27 = −18. And 3abc = 3(1)(2)(−3) = −18.
Both sides match: −18 = −18 ✓
Q2. Find the value of x³+y³−27 if x+y = 3 and xy is unknown... (typical exam phrasing): If x+y−3=0, show x³+y³−27 = 9xy.
Let a=x, b=y, c=−3 so a+b+c=0 ⇒ x³+y³+(−3)³ = 3·x·y·(−3) ⇒ x³+y³−27 = −9xy.
Answer: x³+y³−27 = −9xy
Q3. Factorise a³+b³−6ab+8 (note 8=2³, and think a+b+2... check if a+b+2 relates).
Write as a³+b³+2³−3(a)(b)(2) → factor using the main identity with c=2.
Answer: (a+b+2)(a²+b²+4−ab−2b−2a)

Memory Trick

"Zero sum, triple product." Whenever a question gives you three terms that add up to zero and asks for the sum of their cubes, your alarm bell should ring instantly: the answer is simply 3 times their product.

Common Mistakes

  • Trying to apply the "a+b+c=0" shortcut without checking that the three terms actually sum to zero — always verify first.
  • Forgetting the minus signs in the second bracket — it's a²+b²+c²−ab−bc−ca, all three cross-products subtracted.
  • Confusing this with Identity 8 — Identity 8 squares a trinomial; this one is about cubes and includes an extra −3abc term.

Quick Check: If x+y+z=0, what does x³+y³+z³ equal?

x³+y³+z³ = 3xyz (direct application of the special case)
Identity 10 of 10

Product of Two Binomials

(x+a)(x+b) = x² + (a+b)x + ab

In Plain Language

This identity is the bridge between algebra and quadratic equations. It tells you that whenever you multiply two binomials that share the same variable x, the middle coefficient is the SUM of the two constants, and the last term is their PRODUCT.

Derivation

(x+a)(x+b) = x(x+b) + a(x+b)

= x² + bx + ax + ab

= x² + (a+b)x + ab

Solved Examples

Q1. Expand (x+3)(x+5).
a=3, b=5 → sum=8, product=15.
Answer: x² + 8x + 15
Q2. Expand (x−4)(x+6).
a=−4, b=6 → sum=2, product=−24.
Answer: x² + 2x − 24
Q3. Find 102 × 108 mentally.
Write as (100+2)(100+8) with x=100, a=2, b=8 → 100² + (2+8)(100) + (2)(8) = 10000+1000+16.
Answer: 11016

Memory Trick

"Sum in the middle, product at the end." This is also exactly how you factorise quadratics in reverse — given x²+8x+15, find two numbers that ADD to 8 and MULTIPLY to 15 (here, 3 and 5), and you've cracked it.

Common Mistakes

  • Forgetting signs of a and b when one is negative — always carry the sign into BOTH the sum and the product, e.g. for (x−4)(x+6), a is −4, not 4.
  • Confusing this with Identity 1 — this identity has two DIFFERENT constants a and b, not the same term squared.
  • Adding instead of multiplying for the last term, or vice versa — remember, middle = sum, last = product.

Quick Check: Expand (x−7)(x−2)

a=−7, b=−2 → sum=−9, product=14 → x² − 9x + 14
60-Second Revision

🔄 Flip Cards — Test Yourself Before You Flip

Tap any card to flip it. Try to say the expansion out loud BEFORE you flip — that's how real recall is built, not by re-reading.

(a+b)²
a²+2ab+b²
(a−b)²
a²−2ab+b²
(a+b)(a−b)
a²−b²
(a+b)³
a³+3a²b+3ab²+b³
(a−b)³
a³−3a²b+3ab²−b³
a³+b³
(a+b)(a²−ab+b²)
a³−b³
(a−b)(a²+ab+b²)
(a+b+c)²
a²+b²+c²+2ab+2bc+2ca
a+b+c=0
a³+b³+c³=3abc
Beyond The Exam

🌍 Where These Identities Actually Show Up in Real Life

Algebraic identities aren't just exam tools — they quietly power everyday calculations and entire careers.

🛍️

Fast Mental Maths While Shopping

Calculating a 95-rupee item × 12 quantity, or squaring numbers near 100 (like 98² or 103²), becomes instant using the difference-of-squares and square identities — no calculator needed.

🌾

Land & Farm Area Calculations

If a square plot's side increases by a few metres, (a+b)² instantly tells you the new total area without re-measuring — genuinely useful for farmers and surveyors estimating land.

🏗️

Construction & Engineering

Architects and civil engineers use squared and cubed expressions constantly when calculating areas, volumes of concrete, and load distributions on box-shaped structures.

💻

Computer Science & Cryptography

The difference-of-squares identity is a key trick in number theory used in factorisation algorithms, which underpin parts of modern cryptography and secure communication.

Physics Formulas

Equations like kinetic energy, electrical power (P=I²R), and projectile motion frequently involve squared terms — recognising (a+b)² patterns helps you simplify physics derivations faster.

📈

Higher Studies & Competitive Exams

Every one of these identities reappears in Class 11–12 algebra, JEE/NEET-level problems, and aptitude sections of competitive and government exams — master them now, reuse them for years.

Score Better, Not Just Know More

🎯 Exam Tips From the Examiner's Chair

These small habits are worth real marks in CBSE, SEBA, and other State Board exams.

  • STEP MARKS Always write the identity itself as your first line before substituting values — most marking schemes award a mark just for stating the correct formula, even if a later step has a small slip.
  • SPEED For "find the value of 105² / 998² / 95×105"-type MCQs, identify the nearby round number instantly and apply Identity 1, 2 or 3 — this alone can save 1–2 minutes per paper.
  • SIGNS Before substituting, rewrite every negative term with brackets, e.g., treat (x−5) as a = x, b = −5 rather than risking a sign slip midway.
  • SMART CHOICE When a question gives you a+b and ab (or a−b and ab), always check if it's secretly asking for a²+b², a³+b³, or a³−b³ — these can be solved WITHOUT finding a and b individually.
  • SPECIAL CASE Whenever three terms in a question add to zero, immediately think of a³+b³+c³ = 3abc — this is a favourite HOTS and 3-4 mark question type.
  • VERIFY If time permits, substitute small numbers (like a=1, b=2) into both sides of your final factorised/expanded answer to quickly self-check.
  • PRESENTATION Box or underline your final answer — many State Board evaluators specifically look for a clearly marked final result.
Now Prove It To Yourself

✍️ Practice Question Bank

Attempt each question on paper first, then tap "Show Answer" to check. Resisting the urge to peek early is the single best study habit you can build.

Easy — Direct Application

1. Expand (a + 9)²
a²+18a+81
2. Expand (5 − x)²
25 − 10x + x²
3. Simplify (m + 8)(m − 8)
m² − 64
4. Expand (2a + 3b)²
4a² + 12ab + 9b²
5. Find 999² using a suitable identity
999 = 1000−1 → (1000−1)² = 1000000−2000+1 = 998001

Medium — Apply & Factorise

6. If x + 1/x = 5, find x² + 1/x²
(x+1/x)² = x²+2+1/x² → x²+1/x² = 25 − 2 = 23
7. Factorise 121x² − 144y²
(11x)² − (12y)² = (11x+12y)(11x−12y)
8. Expand (3x − 2y)³
27x³ − 54x²y + 36xy² − 8y³
9. Factorise x³ + 8
x³+2³ = (x+2)(x²−2x+4)
10. Factorise 27a³ − 1
(3a)³−1³ = (3a−1)(9a²+3a+1)
11. If a − b = 5 and ab = 12, find a³ − b³
a³−b³ = (a−b)³+3ab(a−b) = 125 + 3(12)(5) = 125+180 = 305

HOTS — Higher Order Thinking

12. If a+b+c = 15 and a²+b²+c² = 83, find ab+bc+ca
(a+b+c)² = a²+b²+c²+2(ab+bc+ca) → 225 = 83+2(ab+bc+ca) → ab+bc+ca = 71
13. If a = 3, b = −7, and a+b+c = 0, find c, then verify a³+b³+c³ = 3abc
c = 4. a³+b³+c³ = 27−343+64 = −252. 3abc = 3(3)(−7)(4) = −252. Verified ✓
14. Simplify (x+y+z)² − (x²+y²+z²)
2(xy + yz + zx) — the squared terms cancel, leaving only the cross terms.
15. Find the quotient when x³ + 8y³ is divided by (x + 2y), without performing long division
x³+8y³ = (x+2y)(x²−2xy+4y²) → quotient = x² − 2xy + 4y²
16. If x − 1/x = 3, find x³ − 1/x³
x³−1/x³ = (x−1/x)³ + 3(x)(1/x)(x−1/x) = 27 + 3(1)(3) = 27+9 = 36
Before You Go

🌟 Key Takeaways

  • An identity is true for ALL values of the variables, unlike an equation which is true only for specific values.
  • The first three identities — (a+b)², (a−b)², (a+b)(a−b) — are the most exam-frequent and worth memorising until they're instant reflexes.
  • Cube identities follow the 1-3-3-1 pattern from Pascal's Triangle, with alternating signs for the difference form.
  • Factorised forms (a³±b³) always pair an opposite-sign binomial with a same-sign trinomial in the second bracket.
  • The "a+b+c=0" special case is a guaranteed shortcut for a³+b³+c³ = 3abc — examiners love testing this.
  • You rarely need to find individual variable values — most problems can be solved directly using sums, products, and the identities, saving precious exam time.
Frequently Asked

❓ Your Questions, Answered

You absolutely can — the identity gives the exact same result as direct multiplication. The advantage of memorising it is pure speed: in a 3-hour exam with limited time, recognising the pattern and applying the identity in one line is far faster than writing out the full multiplication every single time.
An equation (like x+2=5) holds true for only specific value(s) of the variable. An identity (like (a+b)²=a²+2ab+b²) holds true for every possible value you substitute. That's why identities can be used as universal tools, while equations are solved for particular answers.
Group them in pairs that mirror each other: (a+b)² with (a−b)², a³+b³ with a³−b³, and so on. Notice that within each pair, only the signs change — the structure stays the same. Practising the flip-card revision tool above repeatedly is also far more effective than passive re-reading.
(a+b)², (a−b)², and (a+b)(a−b)=a²−b² appear most frequently across Class 8–10 papers, often in 1-2 mark direct questions. The cube identities and the a³+b³+c³=3abc special case typically appear in 3-4 mark or HOTS-level questions, especially in Class 9 and Class 10.
Yes, completely. These are true identities, which means they hold for any real numbers — positive, negative, fractions, or even algebraic expressions substituted for a and b. Just be extra careful to carry the sign of negative terms correctly through every step.
No — and this is the single most common mistake students make. (a+b)² always equals a²+2ab+b², which includes the extra middle term 2ab. a²+b² is only correct if you're being asked for that expression specifically (not the square of a sum).
Yes. Algebraic identities are universal mathematics — they don't change between boards. CBSE, SEBA, and other State Boards all teach and examine the same core set of identities, usually within Class 8's algebraic expressions chapter and Class 9's polynomials chapter, though exact chapter numbering can vary slightly by textbook edition.
You can re-derive most of them quickly. For example, if you forget (a+b)³, just multiply (a+b)² × (a+b) on rough paper — it only takes about a minute and uses identities you're more likely to remember. This is exactly the derivation method shown for every identity in this guide.
Very directly — factorisation is simply applying these identities in reverse. If you can instantly recognise that an expression matches the pattern a²−2ab+b², you immediately know it factorises to (a−b)². This is why mastering identities now makes the entire factorisation unit in Class 9 and 10 much easier.

You've Got This 💪

Ten identities, one logic. Every formula on this page comes from the same basic idea — multiplying out brackets carefully and combining like terms. There's no magic here, only patterns, and you've now seen exactly where each one comes from. The next time (a+b)² or a³−b³ shows up on your test paper, you won't see a scary jumble of letters — you'll see an old friend.

Keep practising. Keep questioning "why," not just "what." That's how real Maths confidence is built — one identity, one problem, one small win at a time.

📘 Crafted with care by Jnaanangkur – The Learning Hub
Helping NCERT, CBSE & State Board students learn smarter, one concept at a time.

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