Question

8x(cube)+y(cube)-30xy+125, if 2x+y=-5

Answer 1

Here we will be using the identity:

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


We have one piece of info:

$2x+y=-5 \\ \\ \implies \boxed{2x+y+5=0}$


Here we solve it as follows:

$8x^3+y^3+125-30xy \\ \\ \\ = (2x)^3+(y)^3+(5)^3-3(2x)(y)(5) \\ \\ \\ = (2x+y+5)((2x)^2+(y)^2+(5)^2-(2x)(y)-(y)(5)-(5)(2x)) \\ \\ \\ = 0 \times (4x^2+y^2+25-2xy-5y-10x) \\ \\ \\ = 0 \\ \\ \\ \\ \implies \boxed{8x^3+y^3-30xy+125=0}$


Thus, the final answer is 0.


Hope it helps
Purva
Brainly Community

Answer 2

Refer to IMG for answer