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Syllabus
📋 SSC CGL
Subject:Tier-I Examination Syllabus

Algebra & Geometry

Syllabus Unit: Quantitative Aptitude

Why it matters

Chapter overview

This topic pairs the fixed set of algebraic identities used for instant factorisation and simplification with the classical geometry — lines, angles, triangles, circles and mensuration — that SSC CGL tests as direct formula-application MCQs. Together they cover roughly a quarter of the Quantitative Aptitude section, rewarding memorised identities and theorems over derivation from scratch.

Tier-I Examination Syllabus Quantitative Aptitude 25% weightage
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Perfect Square(a±b)² = a²±2ab+b²
Difference of Squaresa²−b²=(a+b)(a−b)
Sum/Diff of Cubesa³±b³ identities
Three-Variable Identitya³+b³+c³−3abc

1Algebraic Identities & Factorisation

A fixed set of identities — (a+b)², a²−b², a³+b³, and a handful more — turns messy factorisation into fast pattern matching.

Learning Objectives

  • Apply (a+b)² = a²+2ab+b², (a−b)² = a²−2ab+b², and a²−b² = (a+b)(a−b) to factorise or simplify expressions.
  • Apply the cube identities a³+b³ = (a+b)(a²−ab+b²) and a³−b³ = (a−b)(a²+ab+b²).
  • Factorise a quadratic trinomial ax²+bx+c by splitting the middle term into two factors of ac summing to b.
  • Apply a³+b³+c³−3abc = (a+b+c)(a²+b²+c²−ab−bc−ca) to three-variable expressions, including the special case where a+b+c=0.

Key Terms

Perfect Square Identity(a±b)² = a² ± 2ab + b².
Difference of Squaresa² − b² = (a+b)(a−b).
Sum/Difference of Cubesa³+b³=(a+b)(a²−ab+b²); a³−b³=(a−b)(a²+ab+b²).
Splitting the Middle TermRewriting bx in ax²+bx+c as two terms whose product equals ac, enabling factorisation by grouping.

Nearly every 'simplify' or 'factorise' question in SSC algebra is a direct match to one of a small set of memorised identities — recognising the pattern is faster than expanding and solving from scratch.

(a+b)² = a²+2ab+b² (a−b)² = a²−2ab+b² a²−b² = (a+b)(a−b)
a³+b³ = (a+b)(a²−ab+b²) a³−b³ = (a−b)(a²+ab+b²)

Worked example: splitting the middle term

x²−5x+6: find two numbers multiplying to +6 and adding to −5 → −2 and −3. So x²−5x+6 = (x−2)(x−3).

Exam TipSpecial case: if a+b+c=0, then a³+b³+c³ = 3abc directly — a frequent SSC shortcut question disguised as a 'find the value' problem.