Question

(X+2b²+c²)/(a+b) +(X+2c²+a²)/(b+c)+(X+2a²+b²)/(c+a) =0;
Find the solution set.

Answer 1

Given (a²-b²)³+(b²-c²)³+(c²-a²)³=3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)
Consider LHS:(a²-b²)³+(b²-c²)³+(c²-a²)³
We know that if x + y + z =0 then x3 + y3 + z3 = 3xyz
Here x = a²-b², y = b²-c² and z = c²-a²
a²-b² + b²-c² + c²-a² = 0
Hence (a²-b²)³+(b²-c²)³+(c²-a²)³ = 3(a²-b²)(b²-c²)(c²-a²)³
= 3(a + b)(a − b)(b + c)(b − c)(c + a)(c − a)
= 3(a + b)(b + c)(c + a)(a − b)(b − c)(c − a)


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