Solve f^-1(y)=(2y+8)^3

$$f ^ { - 1 } ( y ) = ( 2 y + 8 ) ^ { 3 }$$

Solve for f

$$f=\frac{y}{8\left(y+4\right)^{3}}$$

$$y\neq 0\text{ and }y\neq -4$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3} to expand \left(2y+8\right)^{3}.

$$f^{-1}y=8y^{3}+96y^{2}+384y+512$$

Reorder the terms.

$$\frac{1}{f}y=8y^{3}+96y^{2}+384y+512$$

Variable f cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by f.

$$1y=8y^{3}f+96y^{2}f+384yf+f\times 512$$

Swap sides so that all variable terms are on the left hand side.

$$8y^{3}f+96y^{2}f+384yf+f\times 512=1y$$

Reorder the terms.

$$8fy^{3}+96fy^{2}+384fy+512f=y$$

Combine all terms containing f.

$$\left(8y^{3}+96y^{2}+384y+512\right)f=y$$

Divide both sides by 8y^{3}+96y^{2}+384y+512.

$$\frac{\left(8y^{3}+96y^{2}+384y+512\right)f}{8y^{3}+96y^{2}+384y+512}=\frac{y}{8y^{3}+96y^{2}+384y+512}$$

Dividing by 8y^{3}+96y^{2}+384y+512 undoes the multiplication by 8y^{3}+96y^{2}+384y+512.

$$f=\frac{y}{8y^{3}+96y^{2}+384y+512}$$

Divide y by 8y^{3}+96y^{2}+384y+512.

$$f=\frac{y}{8\left(y+4\right)^{3}}$$

Variable f cannot be equal to 0.

$$f=\frac{y}{8\left(y+4\right)^{3}}\text{, }f\neq 0$$