JEE Advanced Maths: Algebra / Polynomials questions with solutions

Q1. Find the sum of all distinct real solutions of the equation (x² - 3)³ - (4x + 6)³ + 216 = 18*(4x + 6)*(3 - x²).

1. -3
2. 4
3. 1
4. -7

Answer: 1

Rewrite equation as (x²-3)³ + (-(4x+6))³ + 6³ - 3*(x²-3)*(-(4x+6))*6 = 0. This factors as (a+b+c)(a²+b²+c²-ab-bc-ca)=0 with a=x²-3, b=-(4x+6), c=6. Factor 1: a+b+c=0 gives x²-4x-3=0, roots x=2+sqrt(7) and x=2-sqrt(7), sum=4. Factor 2: a²+b²+c²-ab-bc-ca=0 gives x=-3. Total sum = 4+(-3) = 1.