Solve 36a^4-97a^2b^2+36b^4

$$36 { a }^{ 4 } -97 { a }^{ 2 } { b }^{ 2 } +36 { b }^{ 4 }$$

Factor

$$\left(2a-3b\right)\left(3a-2b\right)\left(2a+3b\right)\left(3a+2b\right)$$

Solution Steps

Consider 36a^{4}-97a^{2}b^{2}+36b^{4} as a polynomial over variable a.

$$36a^{4}-97b^{2}a^{2}+36b^{4}$$

Find one factor of the form ka^{m}+n, where ka^{m} divides the monomial with the highest power 36a^{4} and n divides the constant factor 36b^{4}. One such factor is 4a^{2}-9b^{2}. Factor the polynomial by dividing it by this factor.

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

Consider 4a^{2}-9b^{2}. Rewrite 4a^{2}-9b^{2} as \left(2a\right)^{2}-\left(3b\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(2a-3b\right)\left(2a+3b\right)$$

Consider 9a^{2}-4b^{2}. Rewrite 9a^{2}-4b^{2} as \left(3a\right)^{2}-\left(2b\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(3a-2b\right)\left(3a+2b\right)$$

Rewrite the complete factored expression.

$$\left(2a-3b\right)\left(2a+3b\right)\left(3a-2b\right)\left(3a+2b\right)$$

Evaluate

$$36a^{4}+36b^{4}-97\left(ab\right)^{2}$$