Question

if a+b+2c=0 ,then prove that a^3+b^3+8c^3=6abc

Answer 1

a+b=-2c
(a+b)^3=(-2c)^3
a^3+b^3+3ab(a+b)=-8c^3
a^3+b^3+{3ab(-2c)}=-8c^3
a^3+b^3+{-6abc)}=-8c^3
a^3+b^3+8c^3=6abc

Answer 2

Heya ✋

Let see your answer !!!!!

Given that

a + b + 2c = 0

We have to prove that

a^3 + b^3 + 8c^3 = 6abc

Solution

a + b + 2c = 0

=> a + b = -2c --------- (i)

On cubing both sides

(a + b)^3 = (-2c)^3

=> a^3 + b^3 + 3ab(a + b) = -8c^3

=> a^3 + b^3 + 3ab(-2c) = -8c^3

=> a^3 + b^3 - 6abc = -8c^3

=> a^3 + b^3 + 8c^3 = 6abc


Proved







Thanks :))))))