Question

Simplify $(xy+yz{)}^2-2x^2{y}^{2}z.$ Find the value when $x=-1,y=1$ and $z=2$.

Answer 1

$(xy+yz{)}^2-2x^2{y}^{2}z$ when $x=-1,y=1,z=2$

$\Rightarrow \left[y\right(x+z){]}^2-2x^2{y}^{2}z$

$\Rightarrow y^2[x^2+z^2+2xz-2x^{2}z]$

$\Rightarrow \left(1{)}^2\right[(-1{)}^2+2^2+2(-1\left)\right(2)-2(-1{)}^2\left(2\right)]$

$\Rightarrow [1+4-4-4]\Rightarrow -3$

$\therefore (xy+yz{)}^2-2x^2{y}^{2}z$ at $x=-1,y=1$ and $z=2$ is $-3$