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I have to simplify this fraction:

$$\frac{a^{-3}+b^{-3}}{a^{-2}-b^{-2}}$$

So I am not sure if I am allowed to do that, but since the exponents were negative, I decided to invert the equation to get positive exponents and then work my way through simplification:

$$\frac{a^{2}-b^{2}}{a^{3}+b^{3}}$$

Then I used difference of squares:

$$\frac{(a-b)(a+b)}{a^{3}+b^{3}}$$

However, I got stuck on this part because I don't know how to simplify $$a^3 + b^3$$

I could simplify it like this:

$$\frac{(a-b)(a+b)}{(a+b)(a^{2}-ab+b^{2})}$$ then cancel out:

$$\frac{(a-b)}{(a^{2}-ab+b^{2})}$$

But wouldn't that still need more simplification?

I am not sure if it can be simplified more or if my inverting of the numerator and denominator are allowed, is it allowed to invert the equation like I did?

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  • $\begingroup$ Your simplification is invalid, you have to multiply by $a^{-2} + b^{-2}$, the conjugate of $a^{-2} + b^{-2}$ on the top and bottom. This method is similar to the method used to evaluate complex numbered fractions $\endgroup$ – Toby Mak Apr 26 '17 at 22:49
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First of all $$\frac{a^{-3}+b^{-3}}{a^{-2}-b^{-2}} \color{red} \neq \frac{a^{2}-b^{2}}{a^{3}+b^{3}}$$

The correct simplification is :$$\frac{a^{-3}+b^{-3}}{a^{-2}-b^{-2}}=\frac{a^{3}+b^{3}}{ab(b^{2}-a^{2})}$$

Now, $$a^3+b^3=(a+b)(a^2-ab+b^2)$$

Therefore :

$$\frac{a^{3}+b^{3}}{ab(b^{2}-a^{2})}=\frac{(a+b)(a^2-ab+b^2)}{ab(b-a)(b+a)}=\color{blue}{\frac{(a^2-ab+b^2)}{ab(b-a)}}$$

That's probably most simplified version of your given expression.

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There is no "simple" reduction of your fraction. Have a look at this.

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  • $\begingroup$ There is a minus sign in the denominator. $\endgroup$ – Donald Splutterwit Apr 26 '17 at 22:52
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    $\begingroup$ Astute observation $\endgroup$ – Allan Apr 26 '17 at 22:52
  • $\begingroup$ Go on Allan ... edit you're answer ... so I can give you +1 too. $\endgroup$ – Donald Splutterwit Apr 26 '17 at 22:57
  • $\begingroup$ Everything in my answer is correct, thanks $\endgroup$ – Allan Apr 26 '17 at 22:58
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    $\begingroup$ The link is correct ... so ... there is a reduction... $\endgroup$ – Donald Splutterwit Apr 26 '17 at 23:03
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$$\frac{\frac{1}{a^3}+\frac{1}{b^3}}{\frac{1}{a^2}-\frac{1}{b^2}}$$ $$\frac{a^3+b^3}{(b^2-a^2)(ab)}$$ $$\frac{a^2+b^2-ab}{(b-a)(ab)}$$

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  • $\begingroup$ Bracket needed in the denominator of the second line. $\endgroup$ – Donald Splutterwit Apr 26 '17 at 22:54
  • $\begingroup$ @DonaldSplutterwit Comment acknowledged $\endgroup$ – The Dead Legend Apr 26 '17 at 22:56
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Your first inversion is incorrect. It is not the case the $a^2-b^2=\frac 1{a^{-2}-b^{-2}}$ The right blows up when $a=b$ while the left is zero, for example. You could do $frac 1{a^{-2}-b^{-2}}=\frac {a^2b^2}{b^2-a^2}$ by multiplying through by $a^2b^2$, but that may not be the most useful thing here.

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You cannot invert the equation because inversion works by multiplying the top and bottom by the variable to a positive exponent.
For example:
$\frac{3}{x^-3} = \frac{3}{\frac{1}{x^3}} * \frac{x^3}{x^3} = \frac{3x^3}{1}$.
You should instead write the expression as nested fractions and follow the rest of the steps like in TheDeadLegend's answer.

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