Simplify this algebraic fraction 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)
 A: 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?
A: Paying attention to the $-$ signs in the second and fourth passage, we get:
$$
\frac{t^4-1}{t^2-t^6}=
\frac{t^4-1}{t^2(1-t^4)}=
\frac{\color{red}{t^4-1}}{-t^2\color{red}{(t^4-1)}}=
\frac{1}{-t^2}=
-\frac{1}{t^2},
$$
for all $t\in\mathbb{R}\backslash\left\{-1,0,1\right\}$.
