Solve 9a^4-18a^2b^2+9b^4

$$9 a ^ { 4 } - 18 a ^ { 2 } b ^ { 2 } + 9 b ^ { 4 }$$

Factor

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

Solution Steps

Factor out 9.

$$9\left(a^{4}-2a^{2}b^{2}+b^{4}\right)$$

Consider a^{4}-2a^{2}b^{2}+b^{4}. Consider a^{4}-2a^{2}b^{2}+b^{4} as a polynomial over variable a.

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

Find one factor of the form a^{k}+m, where a^{k} divides the monomial with the highest power a^{4} and m divides the constant factor b^{4}. One such factor is a^{2}-b^{2}. Factor the polynomial by dividing it by this factor.

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

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).

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

Rewrite the complete factored expression.

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

Evaluate

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