Solve y^4-81x^4

$$y ^ { 4 } - 81 x ^ { 4 }$$

Factor

$$\left(y-3x\right)\left(3x+y\right)\left(9x^{2}+y^{2}\right)$$

Solution Steps

Rewrite y^{4}-81x^{4} as \left(y^{2}\right)^{2}-\left(9x^{2}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(y^{2}-9x^{2}\right)\left(y^{2}+9x^{2}\right)$$

Reorder the terms.

$$\left(-9x^{2}+y^{2}\right)\left(9x^{2}+y^{2}\right)$$

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

$$\left(y-3x\right)\left(y+3x\right)$$

Reorder the terms.

$$\left(-3x+y\right)\left(3x+y\right)$$

Rewrite the complete factored expression.

$$\left(-3x+y\right)\left(3x+y\right)\left(9x^{2}+y^{2}\right)$$

Evaluate

$$y^{4}-81x^{4}$$