# How do you simplify (-7ab^4c)^3[(2a^2c)^2]^3?

Apr 3, 2018

$- 21952 {a}^{15} {b}^{12} {c}^{9}$

#### Explanation:

First of all, let's work with the powers of the round brackets: the power of a product is the power of each single factor

${\left(- 7 a {b}^{4} c\right)}^{3} = {\left(- 7\right)}^{3} \cdot {a}^{3} \cdot {b}^{4 \cdot 3} \cdot {c}^{3} = - 343 {a}^{3} {b}^{12} {c}^{3}$

In the same fashion,

${\left(2 {a}^{2} c\right)}^{2} = {2}^{2} \cdot {a}^{2 \cdot 2} {c}^{2} = 4 {a}^{4} {c}^{2}$

So, we have

$\left(- 343 {a}^{3} {b}^{12} {c}^{3}\right) \cdot {\left(4 {a}^{4} {c}^{2}\right)}^{3}$

Again, we have to cube the last parenthesis:

${\left(4 {a}^{4} {c}^{2}\right)}^{3} = {4}^{3} \cdot {a}^{4 \cdot 3} {c}^{2 \cdot 3} = 64 {a}^{12} {c}^{6}$

Finally, we multiply the two parenthesis:

$\left(- 343 {a}^{3} {b}^{12} {c}^{3}\right) \left(64 {a}^{12} {c}^{6}\right)$

Now we multiply similar factors:

$- 343 \cdot 64 {a}^{3 + 12} {b}^{12} {c}^{3 + 6}$

$- 21952 {a}^{15} {b}^{12} {c}^{9}$