Solve (x-y+1/2)(x-y+1/2)-(x-y)^{2}+(1/3x^{2}-3/2)^3-x^2(frac{1}{27}x^4-frac{1}{2}x^2+frac{9}{4})

$$( x - y + \frac { 1 } { 2 } ) ( x - y + \frac { 1 } { 2 } ) - ( x - y ) ^ { 2 } + ( \frac { 1 } { 3 } x ^ { 2 } - \frac { 3 } { 2 } ) ^ { 3 } - x ^ { 2 } ( \frac { 1 } { 27 } x ^ { 4 } - \frac { 1 } { 2 } x ^ { 2 } + \frac { 9 } { 4 } )$$

Evaluate

$$x-y-\frac{25}{8}$$

Solution Steps

Multiply x-y+\frac{1}{2} and x-y+\frac{1}{2} to get \left(x-y+\frac{1}{2}\right)^{2}.

$$\left(x-y+\frac{1}{2}\right)^{2}-\left(x-y\right)^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Square x-y+\frac{1}{2}.

$$x^{2}-2xy+x+y^{2}-y+\frac{1}{4}-\left(x-y\right)^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x^{2}-2xy+x+y^{2}-y+\frac{1}{4}-\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

To find the opposite of x^{2}-2xy+y^{2}, find the opposite of each term.

$$x^{2}-2xy+x+y^{2}-y+\frac{1}{4}-x^{2}+2xy-y^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$-2xy+x+y^{2}-y+\frac{1}{4}+2xy-y^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Combine -2xy and 2xy to get 0.

$$x+y^{2}-y+\frac{1}{4}-y^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x-y+\frac{1}{4}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x-y+\frac{1}{4}+\frac{1}{27}\left(x^{2}\right)^{3}-\frac{1}{2}\left(x^{2}\right)^{2}+\frac{9}{4}x^{2}-\frac{27}{8}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x-y+\frac{1}{4}+\frac{1}{27}x^{6}-\frac{1}{2}\left(x^{2}\right)^{2}+\frac{9}{4}x^{2}-\frac{27}{8}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x-y+\frac{1}{4}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-\frac{27}{8}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Subtract \frac{27}{8} from \frac{1}{4} to get -\frac{25}{8}.

$$x-y-\frac{25}{8}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Use the distributive property to multiply x^{2} by \frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}.

$$x-y-\frac{25}{8}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-\left(\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}\right)$$

To find the opposite of \frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}, find the opposite of each term.

$$x-y-\frac{25}{8}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-\frac{1}{27}x^{6}+\frac{1}{2}x^{4}-\frac{9}{4}x^{2}$$

Combine \frac{1}{27}x^{6} and -\frac{1}{27}x^{6} to get 0.

$$x-y-\frac{25}{8}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}+\frac{1}{2}x^{4}-\frac{9}{4}x^{2}$$

Combine -\frac{1}{2}x^{4} and \frac{1}{2}x^{4} to get 0.

$$x-y-\frac{25}{8}+\frac{9}{4}x^{2}-\frac{9}{4}x^{2}$$

Combine \frac{9}{4}x^{2} and -\frac{9}{4}x^{2} to get 0.

$$x-y-\frac{25}{8}$$

Expand

$$x-y-\frac{25}{8}$$

Solution Steps

Multiply x-y+\frac{1}{2} and x-y+\frac{1}{2} to get \left(x-y+\frac{1}{2}\right)^{2}.

$$\left(x-y+\frac{1}{2}\right)^{2}-\left(x-y\right)^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Square x-y+\frac{1}{2}.

$$x^{2}-2xy+x+y^{2}-y+\frac{1}{4}-\left(x-y\right)^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x^{2}-2xy+x+y^{2}-y+\frac{1}{4}-\left(x^{2}-2xy+y^{2}\right)+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

To find the opposite of x^{2}-2xy+y^{2}, find the opposite of each term.

$$x^{2}-2xy+x+y^{2}-y+\frac{1}{4}-x^{2}+2xy-y^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$-2xy+x+y^{2}-y+\frac{1}{4}+2xy-y^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Combine -2xy and 2xy to get 0.

$$x+y^{2}-y+\frac{1}{4}-y^{2}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x-y+\frac{1}{4}+\left(\frac{1}{3}x^{2}-\frac{3}{2}\right)^{3}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x-y+\frac{1}{4}+\frac{1}{27}\left(x^{2}\right)^{3}-\frac{1}{2}\left(x^{2}\right)^{2}+\frac{9}{4}x^{2}-\frac{27}{8}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

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

$$x-y+\frac{1}{4}+\frac{1}{27}x^{6}-\frac{1}{2}\left(x^{2}\right)^{2}+\frac{9}{4}x^{2}-\frac{27}{8}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x-y+\frac{1}{4}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-\frac{27}{8}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Subtract \frac{27}{8} from \frac{1}{4} to get -\frac{25}{8}.

$$x-y-\frac{25}{8}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-x^{2}\left(\frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}\right)$$

Use the distributive property to multiply x^{2} by \frac{1}{27}x^{4}-\frac{1}{2}x^{2}+\frac{9}{4}.

$$x-y-\frac{25}{8}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-\left(\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}\right)$$

To find the opposite of \frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}, find the opposite of each term.

$$x-y-\frac{25}{8}+\frac{1}{27}x^{6}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}-\frac{1}{27}x^{6}+\frac{1}{2}x^{4}-\frac{9}{4}x^{2}$$

Combine \frac{1}{27}x^{6} and -\frac{1}{27}x^{6} to get 0.

$$x-y-\frac{25}{8}-\frac{1}{2}x^{4}+\frac{9}{4}x^{2}+\frac{1}{2}x^{4}-\frac{9}{4}x^{2}$$

Combine -\frac{1}{2}x^{4} and \frac{1}{2}x^{4} to get 0.

$$x-y-\frac{25}{8}+\frac{9}{4}x^{2}-\frac{9}{4}x^{2}$$

Combine \frac{9}{4}x^{2} and -\frac{9}{4}x^{2} to get 0.

$$x-y-\frac{25}{8}$$