Solve (2(r-5))/(r-5)^3=4^2V

$$\frac { ( 2 ( r - 5 ) ) } { ( r - 5 ) ^ { 3 } } = 4 ^ { 2 } V$$

Solve for V

$$V=\frac{1}{8\left(r-5\right)^{2}}$$

$$r\neq 5$$

Steps for Solving Linear Equation

Multiply both sides of the equation by \left(r-5\right)^{3}.

$$2\left(r-5\right)=4^{2}V\left(r-5\right)^{3}$$

Use the distributive property to multiply 2 by r-5.

$$2r-10=4^{2}V\left(r-5\right)^{3}$$

Calculate 4 to the power of 2 and get 16.

$$2r-10=16V\left(r-5\right)^{3}$$

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(r-5\right)^{3}.

$$2r-10=16V\left(r^{3}-15r^{2}+75r-125\right)$$

Use the distributive property to multiply 16V by r^{3}-15r^{2}+75r-125.

$$2r-10=16Vr^{3}-240r^{2}V+1200rV-2000V$$

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

$$16Vr^{3}-240r^{2}V+1200rV-2000V=2r-10$$

Combine all terms containing V.

$$\left(16r^{3}-240r^{2}+1200r-2000\right)V=2r-10$$

Divide both sides by 16r^{3}-240r^{2}+1200r-2000.

$$\frac{\left(16r^{3}-240r^{2}+1200r-2000\right)V}{16r^{3}-240r^{2}+1200r-2000}=\frac{2r-10}{16r^{3}-240r^{2}+1200r-2000}$$

Dividing by 16r^{3}-240r^{2}+1200r-2000 undoes the multiplication by 16r^{3}-240r^{2}+1200r-2000.

$$V=\frac{2r-10}{16r^{3}-240r^{2}+1200r-2000}$$

Divide -10+2r by 16r^{3}-240r^{2}+1200r-2000.

$$V=\frac{1}{8\left(r-5\right)^{2}}$$

Solve for r

$$r=5+\frac{\sqrt{\frac{2}{V}}}{4}$$

$$r=5-\frac{\sqrt{\frac{2}{V}}}{4}\text{, }V>0$$