Algebraic Identities for JEE Main: Algebra Deep-Dive

Algebraic identities are the shortcuts that replace laborious computation with instant recognition. JEE Main does not test identities in isolation — they accelerate calculations in quadratic equations, sequences, coordinate geometry, and trigonometry. The students who score highest in algebra are those for whom these identities are reflexive, not looked up. This guide builds that reflexivity for every identity the exam uses.

Core Expansion and Factorisation Identities

(a + b)² = a² + 2ab + b². (a − b)² = a² − 2ab + b². a² − b² = (a+b)(a−b). (a + b)³ = a³ + 3a²b + 3ab² + b³ = a³ + b³ + 3ab(a+b). (a − b)³ = a³ − 3a²b + 3ab² − b³ = a³ − b³ − 3ab(a−b). a³ + b³ = (a+b)(a²−ab+b²). a³ − b³ = (a−b)(a²+ab+b²). These seven are so fundamental that any inability to produce them instantly is costing marks — drill until reflexive. For their use in root expressions see our quadratic equations guide.

The Three-Variable Identity and the a+b+c=0 Corollary

a³ + b³ + c³ − 3abc = (a+b+c)(a²+b²+c²−ab−bc−ca). Corollary: if a + b + c = 0, then a³ + b³ + c³ = 3abc. This corollary solves many seemingly hard problems instantly: if the question sets up three quantities summing to zero, the cube sum is three times their product. Recognising this trigger — three quantities with zero sum — is a high-value pattern recognition skill. JEE uses it in both direct and disguised forms.

Symmetric Functions of Roots

For a quadratic x² − (α+β)x + αβ = 0 with roots α and β: α + β = sum, αβ = product. Key expressions: α² + β² = (α+β)² − 2αβ. α³ + β³ = (α+β)³ − 3αβ(α+β). α² − β² = (α+β)(α−β), where α−β = √((α+β)² − 4αβ). (α + 1/β) + (β + 1/α) = (α+β) + (α+β)/(αβ). These rewrite every symmetric function in terms of sum and product — no need to solve for individual roots. See our sequences and series guide for the symmetric-sum connection to generating functions.

Conditional Identities in Trigonometry

For a triangle where A + B + C = π: sin(A+B) = sinC (and cyclic permutations). Therefore tan A + tan B + tan C = tan A · tan B · tan C. These conditional trig identities follow from applying the algebraic identity (a+b+c = 0 corollary) after converting. Similarly, sin2A + sin2B + sin2C = 4sinA sinB sinC for A+B+C = π. These connect to our properties of triangles guide and appear as direct JEE questions about proving or applying the identity in triangle contexts.

Componendo-Dividendo and Ratio Manipulation

If a/b = c/d, then (a+b)/(a−b) = (c+d)/(c−d). Also, (a+c)/(b+d) = a/b = c/d (addition of equal ratios). These ratio tools simplify equations with fraction variables. The addition-of-equal-ratios result: each ratio also equals (pa + qc)/(pb + qd) for any p, q. Componendo-dividendo is used to simplify an equation like (x+1)/(x−1) = (y+1)/(y−1) — cross-multiply first, then apply it to find x in terms of y efficiently. After reviewing all identities, take a free mock test on algebra to confirm reflexive application.