Solve {l}{(2x-x^{2})^{2}+3x^3(1-x)-2x^2(2-x^2-1/2x)}{(-1/2)^2(2x-y)(2x+y)-(x-1/2y)(x+1/2y)}

$$\left. \begin{array} { l } { ( 2 x - x ^ { 2 } ) ^ { 2 } + 3 x ^ { 3 } ( 1 - x ) - 2 x ^ { 2 } ( 2 - x ^ { 2 } - \frac { 1 } { 2 } x ) } \\ { ( - \frac { 1 } { 2 } ) ^ { 2 } ( 2 x - y ) ( 2 x + y ) - ( x - \frac { 1 } { 2 } y ) ( x + \frac { 1 } { 2 } y ) } \end{array} \right.$$

Sort

$$0,\ 0$$

Solution Steps

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

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

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

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

Use the distributive property to multiply 3x^{3} by 1-x.

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

Combine -4x^{3} and 3x^{3} to get -x^{3}.

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

Combine x^{4} and -3x^{4} to get -2x^{4}.

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

Calculate -\frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

Use the distributive property to multiply \frac{1}{4} by 2x-y.

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

Use the distributive property to multiply \frac{1}{2}x-\frac{1}{4}y by 2x+y and combine like terms.

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

Consider \left(x-\frac{1}{2}y\right)\left(x+\frac{1}{2}y\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(\frac{1}{2}y\right)^{2}.

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

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

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

To find the opposite of x^{2}-\frac{1}{4}y^{2}, find the opposite of each term.

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

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

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

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

$$sort(4x^{2}-x^{3}-2x^{4}-2x^{2}\left(2-x^{2}-\frac{1}{2}x\right),0)$$

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

$$sort(4x^{2}-x^{3}-2x^{4}-4x^{2}+2x^{4}+x^{3},0)$$

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

$$sort(-x^{3}-2x^{4}+2x^{4}+x^{3},0)$$

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

$$sort(-x^{3}+x^{3},0)$$

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

$$sort(0,0)$$

The list values are already in order.

$$0,0$$

Evaluate

$$0,\ 0$$