Solve 4a^8-53a^4b^4+49b^8

$$4 a ^ { 8 } - 53 a ^ { 4 } b ^ { 4 } + 49 b ^ { 8 }$$

Factor

$$\left(a-b\right)\left(a+b\right)\left(2a^{2}-7b^{2}\right)\left(2a^{2}+7b^{2}\right)\left(a^{2}+b^{2}\right)$$

Solution Steps

Consider 4a^{8}-53a^{4}b^{4}+49b^{8} as a polynomial over variable a.

$$4a^{8}-53b^{4}a^{4}+49b^{8}$$

Find one factor of the form ka^{m}+n, where ka^{m} divides the monomial with the highest power 4a^{8} and n divides the constant factor 49b^{8}. One such factor is 4a^{4}-49b^{4}. Factor the polynomial by dividing it by this factor.

$$\left(4a^{4}-49b^{4}\right)\left(a^{4}-b^{4}\right)$$

Consider 4a^{4}-49b^{4}. Rewrite 4a^{4}-49b^{4} as \left(2a^{2}\right)^{2}-\left(7b^{2}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(2a^{2}-7b^{2}\right)\left(2a^{2}+7b^{2}\right)$$

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

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

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.

$$\left(a-b\right)\left(a+b\right)\left(a^{2}+b^{2}\right)\left(2a^{2}-7b^{2}\right)\left(2a^{2}+7b^{2}\right)$$

Evaluate

$$4a^{8}+49b^{8}-53\left(ab\right)^{4}$$