Question

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

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\bf{Solution:-}}}}}}$.

$\large{\underline{\underline{\mathfrak{\bf{Given\:here:-}}}}}$.

• a+b+ 2c = 0.........(1)

$\large{\underline{\underline{\mathfrak{\bf{Proved\:Here:-}}}}}$.

• $\:a^3+b^3+8c^3\:=\:6abc$

$\large{\underline{\underline{\mathfrak{\bf{Explanation:-}}}}}$.

Given Here,

• (a+b+2c)=0

$\implies\:(a+b)\:=\:-2c$

Taking Cube of both side,

$\implies\:(a+b)^3\:=\:(-2c)^3$

$\implies\:a^3+b^3+3ab(a+b)\:=\:(-8c^3)$..........(2)

We know, by equation (1),

• a+b = -2c ..........(3)

Keep value in equation (2) ,

$\implies\:a^3+b^3+3ab(-2c)\:=\:(-8c^3)$

$\boxed{\implies\:a^3+b^3+8c^3\:=\:6abc}$

Thats proved .

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