Solve (x^2)^3-(y^2)^3

$$( x ^ { 2 } ) ^ { 3 } - ( y ^ { 2 } ) ^ { 3 }$$

Evaluate

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

Solution Steps

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$x^{6}-\left(y^{2}\right)^{3}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 3 to get 6.

$$x^{6}-y^{6}$$

Factor

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

Solution Steps

Rewrite x^{6}-y^{6} as \left(x^{3}\right)^{2}-\left(y^{3}\right)^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

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

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

Consider x^{3}+y^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Rewrite the complete factored expression.

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