Solve 16x^3-u^4x^3

$$16 x ^ { 3 } - u ^ { 4 } x ^ { 3 }$$

Factor

$$\left(u-2\right)\left(u+2\right)\left(-u^{2}-4\right)x^{3}$$

Solution Steps

Factor out x^{3}.

$$x^{3}\left(16-u^{4}\right)$$

Consider 16-u^{4}. Rewrite 16-u^{4} as 4^{2}-\left(-u^{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).

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

Reorder the terms.

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

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

$$\left(2-u\right)\left(2+u\right)$$

Reorder the terms.

$$\left(-u+2\right)\left(u+2\right)$$

Rewrite the complete factored expression. Polynomial u^{2}+4 is not factored since it does not have any rational roots.

$$x^{3}\left(u^{2}+4\right)\left(-u+2\right)\left(u+2\right)$$

Evaluate

$$x^{3}\left(16-u^{4}\right)$$