JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The given equation is: (1 - x^3)^ 10 = _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r Using the algebraic identity 1 - x^3 = (1-x)(1+x+x^2), the left hand side can be written as: (1-x)^ 10 (1+x+x^2)^ 10 = _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r Dividing both sides by (1-x)^ 30 : (1-x)^ 10 (1+x+x^2)^ 10 (1-x)^ 30 = _ r=0 ^ 10 a_r x^r (1-x)^ 2r ( 1+x+x^2 (1-x)^2 )^ 10 = _ r=0 ^ 10 a_r ( x (1-x)^2 )^r Let y = x (1-x)^2 . The term inside the bracket on the left hand side can be simplified as: 1+x+x^2 (1-x)^2 = 1-2x+x^2+3x (1-x)^2 = (1-x)^2+3x (1-x)^2 = 1 + 3 ( x (1-x)^2 ) = 1 + 3y Substituting this into the equation, we get: (1 + 3y)^ 10 = _ r=0 ^ 10 a_r y^r Using the binomial expansion, (1 + 3y)^ 10 = _ r=0 ^ 10 ^ 10 C_ r (3y)^r = _ r=0 ^ 10 ^ 10 C_ r 3^r y^r. Comparing the coefficients of y^r on both sides, we obtain: a_r = ^ 10 C_ r 3^r For r = 9 and r = 10: a_9 = ^ 10 C_ 9 3^9 = 10 3^9 a_ 10 = ^ 10 C_ 10 3^ 10 = 1 3^ 10 = 3^ 10 Therefore, the required value is: 9a_9 a_ 10 = 9 10 3^9 3^ 10 = 90 3^9 3 3^9 = 90 3 = 30 Answer: 30