# How do you simplify (3bc)^4?

Nov 11, 2015

$81 {b}^{4} {c}^{4}$

#### Explanation:

What does ${\left(3 b c\right)}^{4}$ mean?

Basically, ${x}^{4} = x \cdot x \cdot x \cdot x$, so in your case, it's

${\left(3 b c\right)}^{4} = \left(3 b c\right) \cdot \left(3 b c\right) \cdot \left(3 b c\right) \cdot \left(3 b c\right)$

As you can multiply in any order your wish, you can drop the parenthesis and "group" the $3$, the $b$ and the $c$ like follows:

${\left(3 b c\right)}^{4} = \left(3 b c\right) \cdot \left(3 b c\right) \cdot \left(3 b c\right) \cdot \left(3 b c\right)$
$= 3 \cdot 3 \cdot 3 \cdot 3 \cdot b \cdot b \cdot b \cdot b \cdot c \cdot c \cdot c \cdot c = {3}^{4} \cdot {b}^{4} \cdot {c}^{4}$

Now, the only thing left to do is computing ${3}^{4}$:

${3}^{4} = {9}^{2} = 81$

So, your solution is $81 {b}^{4} {c}^{4}$.