Solve -81a^4b^4+16c^4

$$- 81 a ^ { 4 } b ^ { 4 } + 16 c ^ { 4 }$$

Factor

$$\left(3ab-2c\right)\left(3ab+2c\right)\left(-9a^{2}b^{2}-4c^{2}\right)$$

Solution Steps

Rewrite -81a^{4}b^{4}+16c^{4} as \left(4c^{2}\right)^{2}-\left(9a^{2}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).

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

Reorder the terms.

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

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

$$\left(2c-3ab\right)\left(2c+3ab\right)$$

Reorder the terms.

$$\left(-3ab+2c\right)\left(3ab+2c\right)$$

Rewrite the complete factored expression.

$$\left(-3ab+2c\right)\left(3ab+2c\right)\left(9a^{2}b^{2}+4c^{2}\right)$$

Evaluate

$$16c^{4}-81\left(ab\right)^{4}$$