Solve (x-1/2y)^3+3(x^2y-1/2y^3)+13/8y^3-3/2xy(x+frac{1}{2}y)

$$( x - \frac { 1 } { 2 } y ) ^ { 3 } + 3 ( x ^ { 2 } y - \frac { 1 } { 2 } y ^ { 3 } ) + \frac { 13 } { 8 } y ^ { 3 } - \frac { 3 } { 2 } x y ( x + \frac { 1 } { 2 } y )$$

Evaluate

$$x^{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-\frac{1}{2}y\right)^{3}.

$$x^{3}-\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{1}{8}y^{3}+3\left(x^{2}y-\frac{1}{2}y^{3}\right)+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Use the distributive property to multiply 3 by x^{2}y-\frac{1}{2}y^{3}.

$$x^{3}-\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{1}{8}y^{3}+3x^{2}y-\frac{3}{2}y^{3}+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Combine -\frac{3}{2}x^{2}y and 3x^{2}y to get \frac{3}{2}x^{2}y.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{1}{8}y^{3}-\frac{3}{2}y^{3}+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Combine -\frac{1}{8}y^{3} and -\frac{3}{2}y^{3} to get -\frac{13}{8}y^{3}.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{13}{8}y^{3}+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Combine -\frac{13}{8}y^{3} and \frac{13}{8}y^{3} to get 0.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Use the distributive property to multiply -\frac{3}{2}xy by x+\frac{1}{2}y.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{3}{2}yx^{2}-\frac{3}{4}xy^{2}$$

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

$$x^{3}+\frac{3}{4}xy^{2}-\frac{3}{4}xy^{2}$$

Combine \frac{3}{4}xy^{2} and -\frac{3}{4}xy^{2} to get 0.

$$x^{3}$$

Expand

$$x^{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(x-\frac{1}{2}y\right)^{3}.

$$x^{3}-\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{1}{8}y^{3}+3\left(x^{2}y-\frac{1}{2}y^{3}\right)+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Use the distributive property to multiply 3 by x^{2}y-\frac{1}{2}y^{3}.

$$x^{3}-\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{1}{8}y^{3}+3x^{2}y-\frac{3}{2}y^{3}+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Combine -\frac{3}{2}x^{2}y and 3x^{2}y to get \frac{3}{2}x^{2}y.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{1}{8}y^{3}-\frac{3}{2}y^{3}+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Combine -\frac{1}{8}y^{3} and -\frac{3}{2}y^{3} to get -\frac{13}{8}y^{3}.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{13}{8}y^{3}+\frac{13}{8}y^{3}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Combine -\frac{13}{8}y^{3} and \frac{13}{8}y^{3} to get 0.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{3}{2}xy\left(x+\frac{1}{2}y\right)$$

Use the distributive property to multiply -\frac{3}{2}xy by x+\frac{1}{2}y.

$$x^{3}+\frac{3}{2}x^{2}y+\frac{3}{4}xy^{2}-\frac{3}{2}yx^{2}-\frac{3}{4}xy^{2}$$

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

$$x^{3}+\frac{3}{4}xy^{2}-\frac{3}{4}xy^{2}$$

Combine \frac{3}{4}xy^{2} and -\frac{3}{4}xy^{2} to get 0.

$$x^{3}$$