Solve w^3x^4-81w^3

$$w ^ { 3 } x ^ { 4 } - 81 w ^ { 3 }$$

Factor

$$\left(x-3\right)\left(x+3\right)\left(x^{2}+9\right)w^{3}$$

Solution Steps

Factor out w^{3}.

$$w^{3}\left(x^{4}-81\right)$$

Consider x^{4}-81. Rewrite x^{4}-81 as \left(x^{2}\right)^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(x-3\right)\left(x+3\right)$$

Rewrite the complete factored expression. Polynomial x^{2}+9 is not factored since it does not have any rational roots.

$$w^{3}\left(x-3\right)\left(x+3\right)\left(x^{2}+9\right)$$

Evaluate

$$w^{3}\left(x^{4}-81\right)$$