# How do you simplify ((3mp^3)/(7q^2))^2?

Nov 15, 2017

${\left(\frac{3 m {p}^{3}}{7 {q}^{2}}\right)}^{2} = \left(\frac{9 {m}^{2} {p}^{6}}{14 {q}^{4}}\right)$

#### Explanation:

${\left(\frac{3 m {p}^{3}}{7 {q}^{2}}\right)}^{2} = \left(\frac{3 m {p}^{3}}{7 {q}^{2}}\right) \cdot \left(\frac{3 m {p}^{3}}{7 {q}^{2}}\right) =$
$\implies \left(\frac{{3}^{2} {m}^{2} {\left({p}^{3}\right)}^{2}}{{\left(7\right)}^{2} {\left({q}^{2}\right)}^{2}}\right)$
$\implies \left(\frac{\left(3 \cdot 3\right) \left(m \cdot m\right) \left({p}^{3} \cdot {p}^{3}\right)}{\left(7 \cdot 7\right) \left({q}^{2} \cdot {q}^{2}\right)}\right)$

Power's laws:
${\left({x}^{a}\right)}^{b} = {x}^{a \cdot b}$
$\left({x}^{a}\right) \left({x}^{b}\right) = {x}^{a + b}$

$\implies \left(\frac{\left(9\right) \left({m}^{2}\right) \left({p}^{3 + 3}\right)}{\left(14\right) \left({q}^{2 + 2}\right)}\right)$
$\implies \left(\frac{\left(9\right) \left({m}^{2}\right) \left({p}^{6}\right)}{\left(14\right) \left({q}^{4}\right)}\right)$