Solve 8w^8-8g^8

$$8 w ^ { 8 } - 8 g ^ { 8 }$$

Factor

$$8\left(w-g\right)\left(w+g\right)\left(w^{2}+g^{2}\right)\left(w^{4}+g^{4}\right)$$

Solution Steps

Factor out 8.

$$8\left(w^{8}-g^{8}\right)$$

Consider w^{8}-g^{8}. Rewrite w^{8}-g^{8} as \left(w^{4}\right)^{2}-\left(g^{4}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(w^{4}-g^{4}\right)\left(w^{4}+g^{4}\right)$$

Consider w^{4}-g^{4}. Rewrite w^{4}-g^{4} as \left(w^{2}\right)^{2}-\left(g^{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).

$$\left(w^{2}-g^{2}\right)\left(w^{2}+g^{2}\right)$$

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

$$\left(w-g\right)\left(w+g\right)$$

Rewrite the complete factored expression.

$$8\left(w-g\right)\left(w+g\right)\left(w^{2}+g^{2}\right)\left(w^{4}+g^{4}\right)$$

Evaluate

$$8\left(w^{8}-g^{8}\right)$$