Solve a^3-9a+a^2-9

$$a ^ { 3 } - 9 a + a ^ { 2 } - 9$$

Factor

$$\left(a-3\right)\left(a+1\right)\left(a+3\right)$$

Solution Steps

Do the grouping a^{3}-9a+a^{2}-9=\left(a^{3}-9a\right)+\left(a^{2}-9\right), and factor out a in a^{3}-9a.

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

Factor out common term a^{2}-9 by using distributive property.

$$\left(a^{2}-9\right)\left(a+1\right)$$

Consider a^{2}-9. Rewrite a^{2}-9 as a^{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).

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

Rewrite the complete factored expression.

$$\left(a-3\right)\left(a+1\right)\left(a+3\right)$$

Evaluate

$$\left(a+1\right)\left(a^{2}-9\right)$$