Solve x^4/16-y^4/81

$$\frac{ { x }^{ 4 } }{ 16 } - \frac{ { y }^{ 4 } }{ 81 }$$

Factor

$$\frac{\left(3x-2y\right)\left(3x+2y\right)\left(9x^{2}+4y^{2}\right)}{1296}$$

Solution Steps

Factor out \frac{1}{1296}.

$$\frac{81x^{4}-16y^{4}}{1296}$$

Consider 81x^{4}-16y^{4}. Rewrite 81x^{4}-16y^{4} as \left(9x^{2}\right)^{2}-\left(4y^{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).

$$\left(9x^{2}-4y^{2}\right)\left(9x^{2}+4y^{2}\right)$$

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

$$\left(3x-2y\right)\left(3x+2y\right)$$

Rewrite the complete factored expression.

$$\frac{\left(3x-2y\right)\left(3x+2y\right)\left(9x^{2}+4y^{2}\right)}{1296}$$

Evaluate

$$\frac{x^{4}}{16}-\frac{y^{4}}{81}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 16 and 81 is 1296. Multiply \frac{x^{4}}{16} times \frac{81}{81}. Multiply \frac{y^{4}}{81} times \frac{16}{16}.

$$\frac{81x^{4}}{1296}-\frac{16y^{4}}{1296}$$

Since \frac{81x^{4}}{1296} and \frac{16y^{4}}{1296} have the same denominator, subtract them by subtracting their numerators.

$$\frac{81x^{4}-16y^{4}}{1296}$$