# How do you simplify (3x^4y^5)^ -3?

Nov 21, 2015

$\frac{1}{27 {x}^{12} {y}^{15}}$

#### Explanation:

Recall that ${\left({x}^{a} {y}^{b} {z}^{c}\right)}^{d} = {x}^{a d} {y}^{b d} {z}^{c d}$. In other words, when something that already has an exponent is being raised to another exponent, you can just multiply the values of the exponents.

So, ${\left(3 {x}^{4} {y}^{5}\right)}^{-} 3 = {3}^{-} 3 {x}^{-} 12 {y}^{-} 15$

Also, recall that ${a}^{-} b = \frac{1}{{a}^{b}}$.

${3}^{-} 3 {x}^{-} 12 {y}^{-} 15 = \frac{1}{{3}^{3} {x}^{12} {y}^{15}} = \frac{1}{27 {x}^{12} {y}^{15}}$