Question

If $x+\frac{1}{x}=5$ and $x^2+\frac{1}{x^3}=8$ then find $x^3+\frac{1}{x^2}$.

Answer 1

Solve for the given expression:

It is given that $x+\frac{1}{x}=5$ and $x^2+\frac{1}{x^3}=8$

On squaring and cubing $x+\frac{1}{x}=5$ on both sides we get

$$\begin{gathered} {\left(x+\frac{1}{x}\right)}^2=25 \\ \Rightarrow x^2+\frac{1}{x^2}+2=25 \\ \Rightarrow x^2+\frac{1}{x^2}=23...\left(i\right) \\ {\left(x+\frac{1}{x}\right)}^3=125 \\ \Rightarrow x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=125 \\ \Rightarrow x^3+\frac{1}{x^3}+3\times 5=125 \\ \Rightarrow x^3+\frac{1}{x^3}=125-15=110...\left(ii\right) \end{gathered}$$

$$\begin{gathered} x^2+\frac{1}{x^2}=23\mathbf{\left[}\mathbf{equation}\mathbf{}\mathbf{\left(}\mathbf{i}\mathbf{\right)}\mathbf{\right]} \\ \Rightarrow 8-\frac{1}{x^3}+\frac{1}{x^2}=23...\left(iii\right)\mathbf{[}\mathbf{\because }{\mathbf{x}}^{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{]} \end{gathered}$$

And

$$\begin{gathered} \frac{1}{x^3}=110-x^3\left[\mathbf{\because }\mathbf{equation}\mathbf{}\mathbf{\left(}\mathbf{ii}\mathbf{\right)}\right] \\ 8+x^3-110+\frac{1}{x^2}=23\left[\mathbf{\because }\mathbf{equation}\mathbf{\left(}\mathbf{iii}\mathbf{\right)}\right] \\ \Rightarrow x^3+\frac{1}{x^2}=125 \end{gathered}$$

Hence the value of $x^3+\frac{1}{x^2}$is $125$.