Solve 18:f(t)=1/(t+1)^3

$$18 : f ( t ) = \frac { 1 } { ( t + 1 ) ^ { 3 } }$$

Solve for f

$$f=18t\left(t+1\right)^{3}$$

$$t\neq -1\text{ and }t\neq 0$$

Solution Steps

Variable f cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by f\left(t+1\right)^{3}, the least common multiple of f,\left(t+1\right)^{3}.

$$\left(t+1\right)^{3}\times 18t=f$$

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

$$\left(t^{3}+3t^{2}+3t+1\right)\times 18t=f$$

Use the distributive property to multiply t^{3}+3t^{2}+3t+1 by 18.

$$\left(18t^{3}+54t^{2}+54t+18\right)t=f$$

Use the distributive property to multiply 18t^{3}+54t^{2}+54t+18 by t.

$$18t^{4}+54t^{3}+54t^{2}+18t=f$$

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

$$f=18t^{4}+54t^{3}+54t^{2}+18t$$

Variable f cannot be equal to 0.

$$f=18t^{4}+54t^{3}+54t^{2}+18t\text{, }f\neq 0$$