Solve a^{4}-a^2b^2-9a^2+9b^2

$$a ^ { 4 } - a ^ { 2 } b ^ { 2 } - 9 a ^ { 2 } + 9 b ^ { 2 }$$

Factor

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

Solution Steps

Do the grouping a^{4}-a^{2}b^{2}-9a^{2}+9b^{2}=\left(a^{4}-a^{2}b^{2}\right)+\left(-9a^{2}+9b^{2}\right), and factor out a^{2} in the first and -9 in the second group.

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

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

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

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

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

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.

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

Evaluate

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