Solve 16x^8/625-256y^4/81

$$\frac{ 16 { x }^{ 8 } }{ 625 } - \frac{ 256 { y }^{ 4 } }{ 81 }$$

Factor

$$\frac{16\left(3x^{2}-10y\right)\left(3x^{2}+10y\right)\left(9x^{4}+100y^{2}\right)}{50625}$$

Solution Steps

Factor out \frac{16}{50625}.

$$\frac{16\left(81x^{8}-10000y^{4}\right)}{50625}$$

Consider 81x^{8}-10000y^{4}. Rewrite 81x^{8}-10000y^{4} as \left(9x^{4}\right)^{2}-\left(100y^{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).

$$\left(9x^{4}-100y^{2}\right)\left(9x^{4}+100y^{2}\right)$$

Consider 9x^{4}-100y^{2}. Rewrite 9x^{4}-100y^{2} as \left(3x^{2}\right)^{2}-\left(10y\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(3x^{2}-10y\right)\left(3x^{2}+10y\right)$$

Rewrite the complete factored expression.

$$\frac{16\left(3x^{2}-10y\right)\left(3x^{2}+10y\right)\left(9x^{4}+100y^{2}\right)}{50625}$$

Evaluate

$$\frac{16x^{8}}{625}-\frac{256y^{4}}{81}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 625 and 81 is 50625. Multiply \frac{16x^{8}}{625} times \frac{81}{81}. Multiply \frac{256y^{4}}{81} times \frac{625}{625}.

$$\frac{81\times 16x^{8}}{50625}-\frac{625\times 256y^{4}}{50625}$$

Since \frac{81\times 16x^{8}}{50625} and \frac{625\times 256y^{4}}{50625} have the same denominator, subtract them by subtracting their numerators.

$$\frac{81\times 16x^{8}-625\times 256y^{4}}{50625}$$

Do the multiplications in 81\times 16x^{8}-625\times 256y^{4}.

$$\frac{1296x^{8}-160000y^{4}}{50625}$$