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Algebra · Topic 1 of 7

🧩 Algebraic Identities

2 exam question types, fully solved

An identity is a formula that is true for every value of the letters. SSC loads its algebra section with them because, once you spot the pattern, a frightening expression collapses in a single line — no slow multiplication needed.

The identities that crack most questions

• Squares: and .

• Cubes: and .

• Cube of a sum: .

The exam habit to build: when you see something like , do not compute — recognise that the top is times the bottom, cancel, and write . Factor first, calculate last.

✨Identity ExplorerExpand & verify on real numbers
a
b
a²−b² with a = 326, b = 222: (a+b)(a−b) = 548 × 104
a²−b² = 56992

Using : .

The next building block is turning a known sum and product into higher powers — that is the sum & product family below: from and you can reach, and even without ever finding or .

📝Practice Questions

Q1Find 53² − 47².

Q2If x + y = 8 and xy = 15, find x² + y².

Q3The value of (87³ + 13³) / (87² − 87×13 + 13²) is:

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Real exam questions — Identities & Sum–Product

2 question types · 8 solved examples from real SSC papers

Algebra in SSC is mostly pattern-spotting: recognise the identity hiding in the question and the long arithmetic disappears. Master these and you answer in seconds, not minutes.

How to solve this type
Three identities crack most of these: a2−b2=(a+b)(a−b)a^2-b^2=(a+b)(a-b), a3−b3=(a−b)(a2+ab+b2)a^3-b^3=(a-b)(a^2+ab+b^2) and a3+b3=(a+b)(a2−ab+b2)a^3+b^3=(a+b)(a^2-ab+b^2). Also recognise the full cube (a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3. The exam trick is to SPOT the pattern and factor — never multiply the big numbers out.

Find the value of 1573−13331572+157×133+1332\dfrac{157^3-133^3}{157^2+157\times 133+133^2}.

A124\dfrac{1}{24}B1290\dfrac{1}{290}C2424D290290

Find 982−2298^2-2^2.

A96009600B98009800C96049604D94009400

If a+b=10a+b=10 and ab=9ab=9, find a3+3a2b+3ab2+b3a^3+3a^2b+3ab^2+b^3.

A100100B900900C729729D10001000

Simplify (a−b)(a+b)(a2+b2)(2a4+2b4)(a-b)(a+b)(a^2+b^2)(2a^4+2b^4).

A2(a8−b8)2(a^8-b^8)Ba8−b82\dfrac{a^8-b^8}{2}Ca4−b4a^4-b^4Da8−b8a^8-b^8