How do you simplify (-3x^2y^-3)^-3/(2x^-2y^-4)?

Aug 12, 2018

$\frac{{y}^{13}}{- 54 {x}^{4}}$

Explanation:

First thing I like to do is convert negative exponents into positive ones.

Remember:
${a}^{- b} = \frac{1}{a} ^ b$

Based on the above, wherever I see a negative exponent, I can move it to the other side of the fraction:

$\implies {\left(- 3 {x}^{2} {y}^{-} 3\right)}^{-} \frac{3}{2 {x}^{-} 2 {y}^{-} 4}$

= (x^2y^4)/(2(-3x^2y^-3)^3

Now, let's go ahead and eliminate that surrounding third power in the denominator, so we can simplify things a bit more easily.

Remember:
${\left({a}^{b}\right)}^{c} = {a}^{b c}$

Bearing this in mind:
=> (x^2y^4)/(2(-3x^2y^-3)^3] = (x^2y^4)/(2(-27x^6y^-9)

Notice that we can now move the ${y}^{-} 9$ up to the numerator and turn it into ${y}^{9}$. We could not move it earlier because we had that third power surrounding the whole expression, but since we eliminated it, now we can:

$\implies \frac{{x}^{2} {y}^{4}}{2 \left(- 27 {x}^{6} {y}^{-} 9\right)} = \frac{{x}^{2} {y}^{4} {y}^{9}}{2 \left(- 27 {x}^{6}\right)}$

Lastly, we eliminate any extraneous terms.

Remember:
${a}^{b} \cdot {a}^{c} = {a}^{b + c}$
${a}^{b} / {a}^{c} = {a}^{b - c}$

Bearing this in mind, we can put together the $y$ terms in the numerator, and cancel out the $x$ terms in the numerator with those in the denominator:

$\implies \frac{{x}^{2} {y}^{4} {y}^{9}}{2 \left(- 27 {x}^{6}\right)} = \frac{{y}^{13}}{2 \left(- 27 {x}^{4}\right)} = \frac{{y}^{13}}{- 54 {x}^{4}}$

As you go through similar problems, be sure to make yourself very comfortable with the highlighted rules. They will be vital to your future math career.

Some useful videos:

Hope that helped :)

Aug 12, 2018

${y}^{13} / \left(- 54 {x}^{4}\right)$

Explanation:

$\frac{{\left(- 3 {x}^{2} {y}^{-} 3\right)}^{-} 3}{2 {x}^{-} 2 {y}^{-} 4}$

$\therefore = \frac{\frac{1}{- 3 {x}^{2} {y}^{-} 3} ^ 3}{2 {x}^{-} 2 {y}^{-} 4}$

$\therefore = \frac{\frac{1}{- {3}^{3} {x}^{6} {y}^{-} 9}}{2 {x}^{-} 2 {y}^{-} 4}$

$\therefore = \frac{1}{- 27 {x}^{6} {y}^{-} 9} \times \frac{1}{2 {x}^{-} 2 {y}^{-} 4}$

$\therefore = \frac{1}{- 54 {x}^{\left(6 - 2\right)} {y}^{\left(- 9 - 4\right)}}$

$\therefore = \frac{1}{- 54 {x}^{4} {y}^{-} 13}$

$\therefore = {y}^{13} / \left(- 54 {x}^{4}\right)$