# How do you simplify (4a^3b^3)(4ab^2)^2 leaving only positive exponents?

Feb 28, 2017

$64 {a}^{5} {b}^{7}$

#### Explanation:

Split it down into easier parts.

${\left(4 a {b}^{2}\right)}^{2}$

Multiply all the inside indices by $2$ because of the $2$ on the outside:

${\left(4 a {b}^{2}\right)}^{2} = {4}^{2} {a}^{2} {b}^{4} = 16 {a}^{2} {b}^{4}$

Now we have

$\left(4 {a}^{3} {b}^{3}\right) \left(16 {a}^{2} {b}^{4}\right)$

You should know that ${a}^{b} \cdot {a}^{c} = {a}^{b + c}$, so

$4 \cdot {a}^{3} \cdot {b}^{3} \cdot 16 \cdot {a}^{2} \cdot {b}^{4}$

$= 4 \cdot 16 \cdot {a}^{3 + 2} \cdot {b}^{3 + 4}$

$= 64 {a}^{5} {b}^{7}$