Solve (2+y)^3-(3y-z)^2

$$( 2 + y ) ^ { 3 } - ( 3 y - z ) ^ { 2 }$$

Evaluate

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

Solution Steps

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

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

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

$$8+12y+6y^{2}+y^{3}-\left(9y^{2}-6yz+z^{2}\right)$$

To find the opposite of 9y^{2}-6yz+z^{2}, find the opposite of each term.

$$8+12y+6y^{2}+y^{3}-9y^{2}+6yz-z^{2}$$

Combine 6y^{2} and -9y^{2} to get -3y^{2}.

$$8+12y-3y^{2}+y^{3}+6yz-z^{2}$$

Expand

$$y^{3}-3y^{2}+6yz+12y-z^{2}+8$$

Solution Steps

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

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

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

$$8+12y+6y^{2}+y^{3}-\left(9y^{2}-6yz+z^{2}\right)$$

To find the opposite of 9y^{2}-6yz+z^{2}, find the opposite of each term.

$$8+12y+6y^{2}+y^{3}-9y^{2}+6yz-z^{2}$$

Combine 6y^{2} and -9y^{2} to get -3y^{2}.

$$8+12y-3y^{2}+y^{3}+6yz-z^{2}$$