How would I solve the following two exponent questions?

(1) The first question is


I got $\quad \displaystyle \frac{-xy^{-2}}{x^2+y},\;\;$but this does not seem to be correct.

(2) My second question is


I got $\quad\displaystyle \frac{A^6B^4}{1/9},\quad$ but my book's answer is $\quad\displaystyle \frac{1}{9A^6B^4}$


We are not solving, we are simplifying: For the first, note that

$$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}\;=\; \frac{xy^2}{x^{-2} + y^{-1}}\;=\;\frac{xy^2}{\large\frac{1}{x^2} + \frac{1}{y}}$$

Try now to multiply numerator and denominator by $x^2y$:

$$\frac{xy^2}{\left(\large\frac{1}{x^2} + \frac{1}{y}\right)}\cdot \frac{(x^2y)}{(x^2y)} \quad = \quad\frac {x\cdot x^2 \cdot y^2 \cdot y}{\left(\large\frac{x^2y}{x^2} + \frac{x^2y}{y}\right)}\quad =\quad \frac{x^3y^3}{y + x^2} $$

For the second, again, we are simplifying:

$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2} \quad = \quad \frac{3^{-2}}{A^{(-3)(-2)}B^{(-2)(-2)}}\quad = \quad \frac{1/9}{A^6B^4}\quad =\quad \frac{1}{9A^6B^4}$$

(as per my now deleted comment below)
Alternatively, for the second problem, we proceed as follows:

$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2}\quad = \quad\left(\frac{A^{-3}B^{-2}}{3}\right)^2 \quad=\quad \frac{A^{-6}B^{-4}}{9} \quad = \quad \frac{1}{9A^6B^4}$$

  • $\begingroup$ One quick question I have on the second problem is dont you flip the fraction when you have a negative exponent. $\endgroup$ – Fernando Martinez Jan 15 '13 at 23:10
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    $\begingroup$ I strongly encourage the OP to check the reason for each step in this answer. If you are unsure of what rule is applied to each step, do some research and/or ask here. $\endgroup$ – Code-Guru Jan 15 '13 at 23:11
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    $\begingroup$ @FernandoMartinez There are often more than one way to simplify these kinds of expressions. $\endgroup$ – Code-Guru Jan 15 '13 at 23:11
  • $\begingroup$ Then is the way I did it initially correct or incorrect? $\endgroup$ – Fernando Martinez Jan 15 '13 at 23:14
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    $\begingroup$ Fernando: you can flip, but you can also multiply through by the negative exponent (since the terms in the numerator and denominator are all multiplied), so each term's exponent can be multiplied by -2, which simplifies the denominator right away. In the first problem, note that at the start, the numerator had a sum, so we can't just multiply through by the exponent. $\endgroup$ – Namaste Jan 15 '13 at 23:14

You got


Here's what I got: $$\left(\frac{x^{-2}+y^{-1}}{xy^2}\right)^{-1}=\left(\frac{xy^2}{x^{-2}+y^{-1}}\right)=\left(\frac{xy^2}{\frac{1}{x^2}+\frac{1}{y}} \right)=\left(\frac{xy^2}{\frac{y+x^2}{yx^2}} \right)=xy^2\cdot\left(\frac{yx^2}{y+x^2} \right)=\frac{x^3y^3}{y+x^2}$$

  • $\begingroup$ Same thing I got, so I think this is the correct answer $\endgroup$ – frogeyedpeas Jan 15 '13 at 22:52

To Solve Problem 1.

We are going to begin by seperating the fraction into two parts (both still underneath one big -1st power)

This leads us to x^-3/y^2 + y^-3/x. After moving the powers to their correct locations we get:

1/(y^2*x^3) + 1(x*y^3).

Now factoring out 1/(x*y^2) from both sides we get.

1/(x*y^2)*[1/(x^2) + 1/(y)]

finding an LCD (least common denominator) for both fractions we can rewrite it as.

(1/x*y^2)*[(y + x^2)/(x^2*y)]

Placing back the factored fraction we end up with:

(y + x^2)/(x^3 * y^3)

and now applying the negative first power (REMEMBER FROM THE START!)

(x^3*y^3)/(y + x^2)

Is the correct answer.

To Solve Number 2.

We begin with the expression (3/(A^(-3) * B^(-2)))^(-2).

First we bring up the A and B (since we can w/ negative exponents!)

so we get (3*A^3 * B^2)^(-2). Now we can evaluate that -2nd power in two steps. First we evaluate it as a 2nd power and then a -1, both we know how to do.

So after evaluating the 2 we end up with:


After the -1 we end up with:


Which is what you should get! Good Luck :)

  • 2
    $\begingroup$ You can find some good starting points on how to format mathematics on the site here and here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. $\endgroup$ – Zev Chonoles Jan 15 '13 at 23:01

Here is another way to simplify the second expression from your question:

$$\left(\frac{3}{A^{-3}B^{-2}}\right)^{-2} = \left(\frac{A^{-3}B^{-2}}{3}\right)^{2} = \left(\frac{A^{-6}B^{-4}}{9}\right) = \left(\frac{1}{9A^{6}B^{4}}\right).$$


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