Solve 1/2left(2y^2+1right)^3-2left(2y^2+1right)left(2y^2-1right)-{left(1/2left(2{y}^2+1right)-1right)}^2-{left(2{y}^3+1/2-yright)}^2

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

Evaluate

$$y^{4}-2y^{3}+3y^{2}+y+2$$

Solution Steps

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

$$\frac{1}{2}\left(8\left(y^{2}\right)^{3}+12\left(y^{2}\right)^{2}+6y^{2}+1\right)-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$\frac{1}{2}\left(8y^{6}+12\left(y^{2}\right)^{2}+6y^{2}+1\right)-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$\frac{1}{2}\left(8y^{6}+12y^{4}+6y^{2}+1\right)-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

Use the distributive property to multiply \frac{1}{2} by 8y^{6}+12y^{4}+6y^{2}+1.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(y^{2}+\frac{1}{2}-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(y^{2}-\frac{1}{2}\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\left(y^{2}\right)^{2}-y^{2}+\frac{1}{4}\right)-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(y^{4}-y^{2}+\frac{1}{4}\right)-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-y^{4}+y^{2}-\frac{1}{4}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

Square 2y^{3}+\frac{1}{2}-y.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-y^{4}+y^{2}-\frac{1}{4}-\left(4y^{6}-4y^{4}+2y^{3}+y^{2}-y+\frac{1}{4}\right)$$

To find the opposite of 4y^{6}-4y^{4}+2y^{3}+y^{2}-y+\frac{1}{4}, find the opposite of each term.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-y^{4}+y^{2}-\frac{1}{4}-4y^{6}+4y^{4}-2y^{3}-y^{2}+y-\frac{1}{4}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)+3y^{4}+y^{2}-\frac{1}{4}-4y^{6}-2y^{3}-y^{2}+y-\frac{1}{4}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)+3y^{4}-\frac{1}{4}-4y^{6}-2y^{3}+y-\frac{1}{4}$$

Subtract \frac{1}{4} from -\frac{1}{4} to get -\frac{1}{2}.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Use the distributive property to multiply -2 by 2y^{2}+1.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}+\left(-4y^{2}-2\right)\left(2y^{2}-1\right)+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Use the distributive property to multiply -4y^{2}-2 by 2y^{2}-1 and combine like terms.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-8y^{4}+2+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Combine 6y^{4} and -8y^{4} to get -2y^{4}.

$$4y^{6}-2y^{4}+3y^{2}+\frac{1}{2}+2+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Add \frac{1}{2} and 2 to get \frac{5}{2}.

$$4y^{6}-2y^{4}+3y^{2}+\frac{5}{2}+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

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

$$4y^{6}+y^{4}+3y^{2}+\frac{5}{2}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Subtract \frac{1}{2} from \frac{5}{2} to get 2.

$$4y^{6}+y^{4}+3y^{2}+2-4y^{6}-2y^{3}+y$$

Combine 4y^{6} and -4y^{6} to get 0.

$$y^{4}+3y^{2}+2-2y^{3}+y$$

Expand

$$y^{4}-2y^{3}+3y^{2}+y+2$$

Solution Steps

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

$$\frac{1}{2}\left(8\left(y^{2}\right)^{3}+12\left(y^{2}\right)^{2}+6y^{2}+1\right)-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$\frac{1}{2}\left(8y^{6}+12\left(y^{2}\right)^{2}+6y^{2}+1\right)-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$\frac{1}{2}\left(8y^{6}+12y^{4}+6y^{2}+1\right)-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

Use the distributive property to multiply \frac{1}{2} by 8y^{6}+12y^{4}+6y^{2}+1.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\frac{1}{2}\left(2y^{2}+1\right)-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(y^{2}+\frac{1}{2}-1\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

Subtract 1 from \frac{1}{2} to get -\frac{1}{2}.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(y^{2}-\frac{1}{2}\right)^{2}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(\left(y^{2}\right)^{2}-y^{2}+\frac{1}{4}\right)-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-\left(y^{4}-y^{2}+\frac{1}{4}\right)-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-y^{4}+y^{2}-\frac{1}{4}-\left(2y^{3}+\frac{1}{2}-y\right)^{2}$$

Square 2y^{3}+\frac{1}{2}-y.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-y^{4}+y^{2}-\frac{1}{4}-\left(4y^{6}-4y^{4}+2y^{3}+y^{2}-y+\frac{1}{4}\right)$$

To find the opposite of 4y^{6}-4y^{4}+2y^{3}+y^{2}-y+\frac{1}{4}, find the opposite of each term.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)-y^{4}+y^{2}-\frac{1}{4}-4y^{6}+4y^{4}-2y^{3}-y^{2}+y-\frac{1}{4}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)+3y^{4}+y^{2}-\frac{1}{4}-4y^{6}-2y^{3}-y^{2}+y-\frac{1}{4}$$

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

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)+3y^{4}-\frac{1}{4}-4y^{6}-2y^{3}+y-\frac{1}{4}$$

Subtract \frac{1}{4} from -\frac{1}{4} to get -\frac{1}{2}.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-2\left(2y^{2}+1\right)\left(2y^{2}-1\right)+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Use the distributive property to multiply -2 by 2y^{2}+1.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}+\left(-4y^{2}-2\right)\left(2y^{2}-1\right)+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Use the distributive property to multiply -4y^{2}-2 by 2y^{2}-1 and combine like terms.

$$4y^{6}+6y^{4}+3y^{2}+\frac{1}{2}-8y^{4}+2+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Combine 6y^{4} and -8y^{4} to get -2y^{4}.

$$4y^{6}-2y^{4}+3y^{2}+\frac{1}{2}+2+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Add \frac{1}{2} and 2 to get \frac{5}{2}.

$$4y^{6}-2y^{4}+3y^{2}+\frac{5}{2}+3y^{4}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

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

$$4y^{6}+y^{4}+3y^{2}+\frac{5}{2}-\frac{1}{2}-4y^{6}-2y^{3}+y$$

Subtract \frac{1}{2} from \frac{5}{2} to get 2.

$$4y^{6}+y^{4}+3y^{2}+2-4y^{6}-2y^{3}+y$$

Combine 4y^{6} and -4y^{6} to get 0.

$$y^{4}+3y^{2}+2-2y^{3}+y$$