Simplify $(x y + y z)^{2} - 2 x^{2} y^{2} z$ and find it's value when $x = - 1, y = 1$ and $z = 2$ .

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-3

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Solution

The correct option is D
-3

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

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

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

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