Question

Simplify:
$\frac{(-2{)}^3\times (3x{)}^2\times (-x{y}^3)}{3x^{2}y}$

Answer 1

$\frac{(-2{)}^3\times (3x{)}^2\times (-x{y}^3)}{3x^{2}y}$
$=\frac{(-2)\times (-2)\times (-2)\times (3{)}^2\times x^2\times (-x)\times y^3}{3\times x^2\times y}$
$=\frac{-8\times 9x^2\times (-1)x\times y^3}{3x^2\times y}$
$=\frac{+72}{3}x{y}^{3-1}$
$=24x{y}^2$