Common Algebraic Identities — The Ones That Keep Showing Up

Algebraic identities are equations that hold true for all values of the variables. Unlike equations (which are true only for specific values), an identity is a mathematical truth. The reason they’re worth memorising is simple: they transform slow, error-prone expansion into instant pattern recognition.

The same identities that appear in Class 8 factorisation reappear in Class 10 polynomials, Class 11 complex numbers, Class 12 limits, and right through JEE Main. Learning them once correctly pays dividends for years.

Key Terms & Definitions

Identity: An algebraic equation that is true for all permissible values of the variables. Example: $(a+b)^2 = a^2 + 2ab + b^2$ — substitute any values of $a$ and $b$ and it holds.

Equation: True only for specific values. Example: $x^2 - 4 = 0$ is true only when $x = \pm 2$ .

Factorisation: Using an identity to rewrite an expression as a product of simpler factors.

LHS/RHS: Left-hand side and right-hand side. When proving an identity, start with one side and transform it to match the other.

The Core Identities You Must Know

Basic Square Identities

(a + b)^2 = a^2 + 2ab + b^2

(a - b)^2 = a^2 - 2ab + b^2

a^2 - b^2 = (a + b)(a - b)

Bonus: $(a+b)^2 - (a-b)^2 = 4ab$

Also: $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$

Three-Variable Identity

(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca

If $a + b + c = 0$ , then: $a^2 + b^2 + c^2 = -2(ab + bc + ca)$

Cube Identities

(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

The Powerful Three-Variable Cube Identity

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

Special case: If $a + b + c = 0$ , then $a^3 + b^3 + c^3 = 3abc$

Understanding the Identities — Not Just Memorising

Why does $(a+b)^2 = a^2 + 2ab + b^2$ ?

Think of a square with side $(a+b)$ . Its area is $(a+b)^2$ . Now cut this square into 4 pieces: a square of side $a$ (area $a^2$ ), a square of side $b$ (area $b^2$ ), and two rectangles of dimensions $a \times b$ (area $2ab$ together). This geometric proof is what CBSE Class 9 actually asks students to “prove” — draw the square and label it.

Why does $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ ?

Expand the RHS: $(a+b)(a^2 - ab + b^2) = a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3 = a^3 + b^3$ . The middle terms cancel out. This is why the factored form has a minus sign on the $ab$ term in the bracket for sum of cubes, and a plus sign for difference of cubes.

The $a+b+c=0$ special case

If $a + b + c = 0$ , then $c = -(a+b)$ . Substituting into the three-variable cube identity:

a^3 + b^3 + (-(a+b))^3 = 3ab(-(a+b))

a^3 + b^3 - (a+b)^3 = -3ab(a+b)

Expanding $(a+b)^3$ and simplifying confirms $a^3 + b^3 + c^3 = 3abc$ .

Solved Examples

Example 1 — CBSE Class 9: Expand $(2x + 3y)^2$

Using $(a+b)^2 = a^2 + 2ab + b^2$ with $a = 2x$ , $b = 3y$ :

(2x + 3y)^2 = (2x)^2 + 2(2x)(3y) + (3y)^2 = 4x^2 + 12xy + 9y^2

Example 2 — CBSE Class 10: Evaluate $103^2 - 97^2$

Using $a^2 - b^2 = (a+b)(a-b)$ with $a = 103$ , $b = 97$ :

103^2 - 97^2 = (103 + 97)(103 - 97) = 200 \times 6 = 1200

This takes 3 seconds. Direct squaring would take 30 seconds and risk arithmetic errors.

Example 3 — Class 9/10: Factorise $8x^3 - 27y^3$

Recognise $8x^3 = (2x)^3$ and $27y^3 = (3y)^3$ . Use $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ :

8x^3 - 27y^3 = (2x - 3y)((2x)^2 + (2x)(3y) + (3y)^2) = (2x - 3y)(4x^2 + 6xy + 9y^2)

Example 4 — JEE Main level: Use the special case

Q: If $a + b + c = 0$ , find the value of $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$ .

Since $a + b + c = 0$ , we know $a^3 + b^3 + c^3 = 3abc$ .

The expression equals $\frac{a^3 + b^3 + c^3}{abc} = \frac{3abc}{abc} = \mathbf{3}$ .

Example 5 — JEE Advanced: Find $a^2 + b^2$ given $a + b$ and $ab$

Q: If $a + b = 5$ and $ab = 6$ , find $a^2 + b^2$ and $a^3 + b^3$ .

$a^2 + b^2 = (a+b)^2 - 2ab = 25 - 12 = 13$

$a^3 + b^3 = (a+b)^3 - 3ab(a+b) = 125 - 3(6)(5) = 125 - 90 = 35$

These “symmetric function” problems appear frequently in JEE — always build up from $(a+b)$ and $ab$ .

Exam-Specific Tips

CBSE Class 9 Polynomials: The $(a+b+c)^2$ expansion and cube factorisation are the favourite 3-mark questions. Also: geometric proof of $(a+b)^2$ identity by area method.

CBSE Class 10: $a^2 - b^2$ is used in simplification questions (like Example 2 above). Students who spot the pattern save enormous time.

JEE Main: The $a + b + c = 0 \Rightarrow a^3 + b^3 + c^3 = 3abc$ trick appears as an MCQ almost every year. Also, symmetric functions (finding $a^n + b^n$ given $a+b$ and $ab$ ) appear in complex numbers and polynomials contexts.

Common Mistakes to Avoid

Mistake 1: Writing $(a + b)^2 = a^2 + b^2$ . This is one of the most common algebra errors in Class 9-10. The middle term $2ab$ is always present. Avoid by always saying aloud “square of first, twice the product, square of second.”

Mistake 2: Wrong sign in $a^3 \pm b^3$ factorisation. For $a^3 + b^3$ , the bracket is $(a^2 - ab + b^2)$ — minus sign on the middle term. For $a^3 - b^3$ , the bracket is $(a^2 + ab + b^2)$ — plus sign. Students mix these up constantly. Memory trick: the sign inside the bracket is opposite to the sign in the original expression.

Mistake 3: Applying $(a + b)^3 = a^3 + b^3$ (missing the three middle terms). The full expansion is $a^3 + 3a^2b + 3ab^2 + b^3$ . Note the coefficients follow Pascal’s triangle: 1, 3, 3, 1.

Mistake 4: Using $a^2 - b^2$ as a single “identity” but forgetting both factors — writing $(a+b)$ only and forgetting the $(a-b)$ factor. Always: $a^2 - b^2 = (a+b)(a-b)$ , both factors.

Mistake 5: In the identity $(a + b + c)^2$ , forgetting that the cross-product terms have coefficient 2: $2ab + 2bc + 2ca$ . Students often write $ab + bc + ca$ (coefficient 1).

Practice Questions

Q1. Using an identity, evaluate $998^2$ .

$998^2 = (1000 - 2)^2 = 1000^2 - 2(1000)(2) + 2^2 = 1000000 - 4000 + 4 = 996004$ .

Q2. Factorise $49x^2 - 25y^2$ .

$(7x)^2 - (5y)^2 = (7x + 5y)(7x - 5y)$ .

Q3. If $x + y = 4$ and $xy = 3$ , find $x^2 + y^2$ and $x^3 + y^3$ .

$x^2 + y^2 = (x+y)^2 - 2xy = 16 - 6 = 10$ . $x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 64 - 36 = 28$ .

Q4. Prove that if $a + b + c = 7$ and $ab + bc + ca = 20$ , then $a^2 + b^2 + c^2 = 9$ .

$(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$ . So $49 = a^2+b^2+c^2 + 40$ . Therefore $a^2+b^2+c^2 = 9$ .

Q5. Factorise $x^3 + 8$ .

$x^3 + 8 = x^3 + 2^3 = (x+2)(x^2 - 2x + 4)$ .

Q6. Find the value of $125x^3 + 27y^3$ given $5x + 3y = 2$ and $xy = 4$ .

$125x^3 + 27y^3 = (5x+3y)^3 - 3(5x)(3y)(5x+3y) = 8 - 3(15xy)(2) = 8 - 90xy = 8 - 360 = -352$ .

Q7. If $a - b = 4$ and $a^2 + b^2 = 40$ , find $ab$ .

$(a-b)^2 = a^2 - 2ab + b^2$ . So $16 = 40 - 2ab$ , giving $2ab = 24$ , $ab = 12$ .

Q8. Show that $(a + b - c)^2 + (b + c - a)^2 + (c + a - b)^2 = 3(a^2 + b^2 + c^2) - 2(ab + bc + ca) - (a^2 + b^2 + c^2)$ … [verify the simplified form is $a^2 + b^2 + c^2 + 2ab + 2bc + 2ca - 4ab - 4bc - 4ca = ...$ ]

Expand each: $(a+b-c)^2 = a^2+b^2+c^2+2ab-2bc-2ca$ . Summing all three gives $3a^2+3b^2+3c^2 - 2ab - 2bc - 2ca = 3(a^2+b^2+c^2) - 2(ab+bc+ca)$ .

Advanced Identity Applications

Sophie Germain Identity (JEE Advanced)

a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab)

This identity is used to factorise expressions that look like “sum of two fourth powers” — which cannot be factored using the standard $a^2 + b^2$ (since $a^2 + b^2$ is irreducible over reals).

Application: Show that $n^4 + 4$ is composite for all integers $n > 1$ .

Using Sophie Germain with $a = n$ , $b = 1$ : $n^4 + 4 = (n^2 + 2 + 2n)(n^2 + 2 - 2n) = (n^2 + 2n + 2)(n^2 - 2n + 2)$

For $n > 1$ : both factors are greater than 1 (check: $n^2 - 2n + 2 = (n-1)^2 + 1 \geq 2$ ). So $n^4 + 4$ is always the product of two integers both greater than 1 — hence composite.

Newton’s Identities — Sums of Powers from Symmetric Functions

If $\alpha$ and $\beta$ are roots of $x^2 - px + q = 0$ (so $\alpha + \beta = p$ , $\alpha\beta = q$ ), then:

\alpha^n + \beta^n = p(\alpha^{n-1} + \beta^{n-1}) - q(\alpha^{n-2} + \beta^{n-2})

This recursion lets you find $\alpha^4 + \beta^4$ , $\alpha^5 + \beta^5$ , etc. without knowing $\alpha$ and $\beta$ individually.

Example: If $\alpha + \beta = 3$ and $\alpha\beta = 2$ , find $\alpha^4 + \beta^4$ .

$\alpha^2 + \beta^2 = p^2 - 2q = 9 - 4 = 5$

$\alpha^3 + \beta^3 = p(\alpha^2 + \beta^2) - q(\alpha + \beta) = 3(5) - 2(3) = 9$

$\alpha^4 + \beta^4 = p(\alpha^3 + \beta^3) - q(\alpha^2 + \beta^2) = 3(9) - 2(5) = 17$

JEE Main 2024 asked: “If $x + 1/x = 3$ , find $x^4 + 1/x^4$ .” Strategy: $x^2 + 1/x^2 = (x + 1/x)^2 - 2 = 7$ . Then $x^4 + 1/x^4 = (x^2 + 1/x^2)^2 - 2 = 49 - 2 = 47$ . This repeated squaring pattern uses the identity $(a+b)^2 = a^2 + 2ab + b^2$ twice.

Q9. If $a + b + c = 0$ , find $\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab}$ .

$\frac{a^2}{bc} + \frac{b^2}{ca} + \frac{c^2}{ab} = \frac{a^3 + b^3 + c^3}{abc}$ . Since $a + b + c = 0$ , we know $a^3 + b^3 + c^3 = 3abc$ . Therefore the expression $= \frac{3abc}{abc} = \mathbf{3}$ .

FAQs

Why do we need identities — can’t we just expand everything?

Expanding works but is slow and error-prone for large numbers or complex expressions. Identities let us do reverse work (factorisation) and simplification instantly. In competitive exams like JEE, where speed matters enormously, pattern recognition through identities can save 1–2 minutes per question.

Are there identities beyond Class 12 level worth knowing?

The Binomial Theorem $(a+b)^n$ for any $n$ generalises all the cube and square identities. At JEE level, knowing the first few terms of the binomial expansion is essential for approximation problems (e.g., finding $1.02^{10}$ approximately).

How do I remember which sign goes inside the bracket for cube factorisation?

Sum of cubes: $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ . The sign inside the trinomial bracket is opposite to the sign in the original. Difference of cubes: $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ . Same rule applies.

Can $(a+b)^2 = a^2 + b^2$ ever be true?

Only when $ab = 0$ , i.e., when $a = 0$ or $b = 0$ . In general, for non-zero $a$ and $b$ , the middle term $2ab$ is never zero and cannot be ignored.