Question

Find the value of $3x^2\times \left(\frac{-1}{9}x{y}^2\right)\times 4xy{z}^2$, when $x=1,y=2$ and $z=3$.

Answer 1

$3x^2\times \frac{-x{y}^2}{9}\times 4xy{z}^2$

$=\frac{-4x^4{y}^3{z}^2}{3}$ when $x=1,y=2,z=3$

$=\frac{-4\times 1^4\times 2^3\times 3^2}{3}$

$=-4\times 1^4\times 2^3\times 3^{2-1}$ since $\frac{a^m}{a^n}=a^{m-n}$

$=-4\times 1^4\times 2^3\times 3$

$=-4\times 1\times 8\times 3=-96$