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I have this algebraic fraction:

$$\frac{t^4-1}{t^2-t^6}$$

And I'm told the answer is:

$$\frac{-1}{t^2}$$

I can't for the life of me work out how to simplify it. (I'm sorry for the simple question)

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3  
Can you see that the bottom is $t^2(1-t^4)$? –  André Nicolas May 31 '12 at 0:26

1 Answer 1

When faced with the problem of simplifying $$\frac{x^4-1}{x^2-x^6}\;,$$ you have almost too many possibilities. For instance, $x^4-1=(x^2)^2-1^2$ is a difference of squares, something that’s often useful in simplification. But one thing that should leap out at you is that $x^2$ is a common factor of both terms in the denominator, so we can write

$$\frac{x^4-1}{x^2-x^6}=\frac{x^4-1}{x^2(1-x^4)}\;.$$

If that had only been $x^2(x^4-1)$ instead, we could do the obvious cancellation and get $$\frac{x^4-1}{x^2(x^4-1)}=\frac1{x^2}\cdot\frac{x^4-1}{x^4-1}=\frac1{x^2}\;.\tag{1}$$ It isn’t but it’s close, and the desired answer is close to $\dfrac1{x^2}$, so we might try to imitate $(1)$ as far as possible:

$$\frac{x^4-1}{x^2(1-x^4)}=\frac1{x^2}\cdot\frac{x^4-1}{1-x^4}=\stackrel{???}\dots=-\frac1{x^2}\;.$$

Can you finish it from there?

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+1; everything which could be said was said. –  000 May 31 '12 at 2:41
    
I managed to get up to the question marks but then I am stuck again, could you factories -1 out of the denominator? –  Jonathan. May 31 '12 at 9:47
    
@Jonathan: You could indeed: $1-x^4=-(x^4-1)$. You could equally well factor it out of the numerator: $x^4-1=-(1-x^4)$. –  Brian M. Scott May 31 '12 at 9:49

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