Solve 8b^4-648

$$8 b ^ { 4 } - 648$$

Factor

$$8\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)$$

Solution Steps

Factor out 8.

$$8\left(b^{4}-81\right)$$

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

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

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

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

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

$$8\left(b-3\right)\left(b+3\right)\left(b^{2}+9\right)$$

Evaluate

$$8\left(b^{4}-81\right)$$