Solve 16a^4-8a^2b^2+b^4

$$16 a ^ { 4 } - 8 a ^ { 2 } b ^ { 2 } + b ^ { 4 }$$

Factor

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

Solution Steps

Consider 16a^{4}-8a^{2}b^{2}+b^{4} as a polynomial over variable a.

$$16a^{4}-8b^{2}a^{2}+b^{4}$$

Find one factor of the form ka^{m}+n, where ka^{m} divides the monomial with the highest power 16a^{4} and n divides the constant factor b^{4}. One such factor is 4a^{2}-b^{2}. Factor the polynomial by dividing it by this factor.

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

Consider 4a^{2}-b^{2}. Rewrite 4a^{2}-b^{2} as \left(2a\right)^{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).

$$\left(2a-b\right)\left(2a+b\right)$$

Rewrite the complete factored expression.

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

Evaluate

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