Solve left(x+yright)^3-3xyleft(x+yright)^

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

Evaluate

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

Solution Steps

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

$$x^{3}+3x^{2}y+3xy^{2}+y^{3}-3xy\left(x+y\right)^{1}$$

Calculate x+y to the power of 1 and get x+y.

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

Use the distributive property to multiply -3xy by x+y.

$$x^{3}+3x^{2}y+3xy^{2}+y^{3}-3yx^{2}-3xy^{2}$$

Combine 3x^{2}y and -3yx^{2} to get 0.

$$x^{3}+3xy^{2}+y^{3}-3xy^{2}$$

Combine 3xy^{2} and -3xy^{2} to get 0.

$$x^{3}+y^{3}$$

Expand

$$x^{3}+y^{3}$$

Solution Steps

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

$$x^{3}+3x^{2}y+3xy^{2}+y^{3}-3xy\left(x+y\right)^{1}$$

Calculate x+y to the power of 1 and get x+y.

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

Use the distributive property to multiply -3xy by x+y.

$$x^{3}+3x^{2}y+3xy^{2}+y^{3}-3yx^{2}-3xy^{2}$$

Combine 3x^{2}y and -3yx^{2} to get 0.

$$x^{3}+3xy^{2}+y^{3}-3xy^{2}$$

Combine 3xy^{2} and -3xy^{2} to get 0.

$$x^{3}+y^{3}$$