Question

a+b+2c=0 proove that a^3+b^3+8c^3=6abc

Answer 1

In the attachment I have answered this problem.

I have applied the algebraic identity

(a+b)^3 = a^3 +b^3 + 3ab(a+b)

See the attachment for detailed solution

Answer 2

Step-by-step explanation:

Given:

• a + b + 2c = 0

To prove:

• a³ + b³ + 8c³ = 6abc

Solution:

$\small\implies{\sf }$ a+b+2c= 0

$\small\implies{\sf }$ a+b = –2c ..........(1)

$\small\implies{\sf }$ Cubing on both the sides, We have

$\small\implies{\sf }$ (a+b)³ =(–2c)³

We know that (a+b)³ = a³+b³+3ab(a+b)

$\small\implies{\sf }$ a³+b³+3ab(a+b) = –8c³

$\small\implies{\sf }$ a³+b³+3ab(–2c) = –8c³ (∴ a+b = –2c)

→ a³+b³–6abc = – 8c³

$\small\implies{\sf }$ a³+b³+8c³ = 6abc

Hence, L.H.S = R.H.S