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

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

Evaluate

$$0$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{3}-y^{3}\right)^{2}.

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

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

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

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

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

Combine x and -x to get 0.

$$\left(x^{6}-2x^{3}y^{3}+y^{6}\right)\times 0$$

Anything times zero gives zero.

$$0$$

Factor

$$0$$