Solve (-x)^2(-x)^3=(-x)^5=-x^5

$$( - x ) ^ { 2 } ( - x ) ^ { 3 } = ( - x ) ^ { 5 } = - x ^ { 5 }$$

Evaluate (complex solution)

Solution Steps

To multiply powers of the same base, add their exponents. Add 2 and 3 to get 5.

$$\left(-x\right)^{5}=\left(-x\right)^{5}\text{ and }\left(-x\right)^{5}=-x^{5}$$

Subtract \left(-x\right)^{5} from both sides.

$$\left(-x\right)^{5}-\left(-x\right)^{5}=0\text{ and }\left(-x\right)^{5}=-x^{5}$$

Combine \left(-x\right)^{5} and -\left(-x\right)^{5} to get 0.

$$0=0\text{ and }\left(-x\right)^{5}=-x^{5}$$

Compare 0 and 0.

$$\text{true}\text{ and }\left(-x\right)^{5}=-x^{5}$$

Add x^{5} to both sides.

$$\text{true}\text{ and }\left(-x\right)^{5}+x^{5}=0$$

Expand \left(-x\right)^{5}.

$$\text{true}\text{ and }\left(-1\right)^{5}x^{5}+x^{5}=0$$

Calculate -1 to the power of 5 and get -1.

$$\text{true}\text{ and }-x^{5}+x^{5}=0$$

Combine -x^{5} and x^{5} to get 0.

$$\text{true}\text{ and }0=0$$

Compare 0 and 0.

$$\text{true}\text{ and }\text{true}$$

The conjunction of \text{true} and \text{true} is \text{true}.

$$\text{true}$$

Solve for x

$$x\in \mathrm{R}$$