Question

if a+b-2c=0, find the value of a3+b3-8c3+6abc​

Answer 1

Answer:

0

Step-by-step explanation:

We are given $\sf{a^3+b^3-8b^3+6abc}$.

From its structure, we can tell

$\sf{(a)^3+(b)^3+(-2c)^3-3\time(a)(b)(-2c)}$

From the identity, we obtain

$\sf{(a+b-2c)(a^2+b^2+4c^2-ab+2bc+2ca)}$

Because of $\sf{a+b-2c=0}$

$\sf{0\times(a^2+b^2+4c^2-ab+2bc+2ca)=0}$

For your information.

1. The identity used above is

$\sf{x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)}$

2. Why do we use $\sf{a+b-2c=0}$?

Since multiplication by zero gives zero.

The result is 0.