Simplify 8(2t(1 - t^2) - 2t(-t^2 - 1)) / (1 - t^2)^2 - 10(1 - t^2) / (t^2 + 1)^2.

Step-by-step solution

1. 1

   Clean the inner bracket before anything else. The numerator of the first fraction hides a large cancellation; expanding it first shrinks the problem dramatically:

   $2t\left(1-t^2\right)=2t-2t^3,\qquad 2t\left(-t^2-1\right)=-2t^3-2t$

2. 2

   Subtract, distributing the minus over both terms.

   $\left(2t-2t^3\right)-\left(-2t^3-2t\right)=2t-2t^3+2t^3+2t=4t$

   The cubic terms cancel entirely — the bracket is just $4t$.

3. 3

   Rebuild the first fraction. Multiplying by the outer $8$:

   $\frac{8\cdot 4t}{\left(1-t^2\right)^2}=\frac{32t}{\left(1-t^2\right)^2}$

   so the expression is now

   $\frac{32t}{\left(1-t^2\right)^2}-\frac{10\left(1-t^2\right)}{\left(1+t^2\right)^2}$

4. 4

   Check whether the two denominators share a factor. $\left(1-t^2\right)^2=(1-t)^2(1+t)^2$ while $\left(1+t^2\right)^2$ is irreducible over the reals. They have no common factor, so the least common denominator is simply their product:

   $\left(1-t^2\right)^2\left(1+t^2\right)^2$

5. 5

   Write over the common denominator.

   $\frac{32t\left(1+t^2\right)^2-10\left(1-t^2\right)^3}{\left(1-t^2\right)^2\left(1+t^2\right)^2}$

   The numerator does not factor against the denominator, so this is as far as the simplification goes; the two-fraction form is often the more usable one.

6. 6

   Verify numerically. At $t=2.1$ the original expression, the two-fraction form and the single-fraction form all equal $6.9441950704$; at $t=-1.4$ all three give $-47.5154208262$. $\checkmark$ (Both forms are undefined at $t=\pm 1$.)