Question

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

• A. 3
• B. 4
• C. -3
• D. 0

Answer 1

The correct option is C

-3

Simplify ${(xy+yz)}^2$ using the identity ${(a+b)}^2$ $=$ $a^2$+ $b^2$ + 2ab
We get, ${(xy+yz)}^2$=$x^2$$y^2$ $+$ $y^2$$z^2$ $+$ ${2xzy}^2$
Therefore, ${(xy+yz)}^2$$-$$2x^2$$y^2$z $=$ $x^2$$y^2$ $+$ $y^2$$z^2$ $+$ 2xz$y^2$ $-$ 2$x^2$$y^2$z
Placing the values of x,y and z in the above expression,
$1+4-4-4=-3$