# How do you simplify sqrt(4c^3d^3)*sqrt(8c^3d)?

Apr 30, 2017

$4 \sqrt{2} {c}^{3} {d}^{2}$

#### Explanation:

Supposing that everything is well defined:

$\sqrt{4 {c}^{3} {d}^{3}} \sqrt{8 {c}^{3} d} = \sqrt{{2}^{2 + 3} {c}^{3 + 3} {d}^{3 + 1}} = {2}^{\frac{5}{2}} {c}^{\frac{6}{2}} {d}^{\frac{4}{2}} = 4 \sqrt{2} {c}^{3} {d}^{2}$

Apr 30, 2017

See the solution process below:

#### Explanation:

We can use this rule for multiplying radicals to rewrite the expression:

$\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$

$\sqrt{4 {c}^{3} {d}^{3}} \cdot \sqrt{8 {c}^{3} d} = \sqrt{4 {c}^{3} {d}^{3} \cdot 8 {c}^{3} d} = \sqrt{32 {c}^{3} {c}^{3} {d}^{3} d}$

We can use these rules for exponents to rewrite the $c$ and $d$ terms:

$a = {a}^{\textcolor{red}{1}}$ and ${x}^{\textcolor{red}{a}} \times {x}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} + \textcolor{b l u e}{b}}$

$\sqrt{32 {c}^{3} {c}^{3} {d}^{3} d} = \sqrt{32 {c}^{\textcolor{red}{3}} {c}^{\textcolor{b l u e}{3}} {d}^{\textcolor{red}{3}} {d}^{\textcolor{b l u e}{1}}} = \sqrt{32 {c}^{\textcolor{red}{3} + \textcolor{b l u e}{3}} {d}^{\textcolor{red}{3} + \textcolor{b l u e}{1}}} = \sqrt{32 {c}^{6} {d}^{4}}$

We can now rewrite the expression and take the square root of terms:

$\sqrt{32 {c}^{6} {d}^{4}} = \sqrt{2 \cdot \left(16 {c}^{6} {d}^{4}\right)} = \sqrt{2} \cdot \sqrt{16 {c}^{6} {d}^{4}} = \sqrt{2} \cdot \pm 4 {c}^{3} {d}^{2} = \pm 4 \sqrt{2} {c}^{3} {d}^{2}$