Question

Simplify $(xy+yz{)}^2-2x^2{y}^{2}z$ and find it's 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

${(xy+yz)}^2$ is in the form of ${(a+b)}^2$ where $a=xy$ and $b=yz.$
Using ${(a+b)}^2=a^2+b^2+2ab,$
${(xy+yz)}^2=x^2{y}^2+y^2{z}^2+{2xzy}^2.$

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

Substituting the values of $x$,$y$ and $z$ in the above expression, we get
$x^2{y}^2+y^2{z}^2+2xz{y}^2-2x^2{y}^{2}z$
${(-1)}^2{\left(1\right)}^2+{\left(1\right)}^2{\left(2\right)}^2+2(-1)\left(2\right){\left(1\right)}^2-2{\left(1\right)}^2{\left(1\right)}^2\left(2\right)$
1 + 4 - 4 - 4 = -3