Solve -left(x^2right)^3+x^2

$$- { \left( { x }^{ 2 } \right) }^{ 3 } + { x }^{ 2 }$$

Factor

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

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$factor(-x^{6}+x^{2})$$

Factor out x^{2}.

$$x^{2}\left(-x^{4}+1\right)$$

Consider -x^{4}+1. Rewrite -x^{4}+1 as 1^{2}-\left(-x^{2}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Reorder the terms.

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

Consider -x^{2}+1. Rewrite -x^{2}+1 as 1^{2}-x^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(1-x\right)\left(1+x\right)$$

Reorder the terms.

$$\left(-x+1\right)\left(x+1\right)$$

Rewrite the complete factored expression. Polynomial x^{2}+1 is not factored since it does not have any rational roots.

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

Evaluate

$$x^{2}-x^{6}$$

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$-x^{6}+x^{2}$$