How do you divide (-3x^2w)^3/(3x^4w^2)^4?

Jul 25, 2015

You use three properties of exponents to rewrite the numerator and denominator, and cancel out common terms.

Explanation:

Three properties of exponents will come in handy for this problem

• power of a power property

${\left({x}^{n}\right)}^{m} = {x}^{n \cdot m}$

• power of a product property

${\left(x \cdot y\right)}^{n} = {x}^{2} \cdot {y}^{n}$

• quotient of powers property

$\frac{{x}^{n}}{{x}^{m}} = {x}^{n - m}$, where $x \ne 0$

Using these three properties will allow you to rewrite the numerator and denominator as

${\left(- 3 {x}^{2} w\right)}^{3} = {\left(- 3\right)}^{3} \cdot {\left({x}^{2}\right)}^{3} \cdot {w}^{3} = - 27 \cdot {x}^{6} \cdot {w}^{3}$

and

${\left(3 {x}^{4} {w}^{2}\right)}^{4} = {3}^{4} \cdot {\left({x}^{4}\right)}^{4} \cdot {\left({w}^{2}\right)}^{4} = 81 \cdot {x}^{16} \cdot {w}^{8}$

The expression will now become

$\frac{- \cancel{27} \cdot {x}^{6} \cdot {w}^{3}}{\cancel{27} \cdot 3 \cdot {x}^{16} \cdot {w}^{8}} = - \frac{{x}^{6} \cdot {w}^{3}}{3 \cdot {x}^{16} \cdot {w}^{8}}$

Finally, the expression becomes

$- \frac{{x}^{6} \cdot {w}^{3}}{3 \cdot {x}^{16} \cdot {w}^{8}} = \textcolor{g r e e n}{- \frac{1}{3 \cdot {x}^{10} \cdot {w}^{5}}}$