Solve (t^2-1)^2-9

$$( t ^ { 2 } - 1 ) ^ { 2 } - 9$$

Evaluate

$$t^{4}-2t^{2}-8$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t^{2}-1\right)^{2}.

$$\left(t^{2}\right)^{2}-2t^{2}+1-9$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$t^{4}-2t^{2}+1-9$$

Subtract 9 from 1 to get -8.

$$t^{4}-2t^{2}-8$$

Expand

$$t^{4}-2t^{2}-8$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t^{2}-1\right)^{2}.

$$\left(t^{2}\right)^{2}-2t^{2}+1-9$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$t^{4}-2t^{2}+1-9$$

Subtract 9 from 1 to get -8.

$$t^{4}-2t^{2}-8$$