Question

Cube Root of [3 {(cube root of x) - (1/cube root of x)}] = 2. Find x .
After finding 'x' what is the value of [(cube root of x ) + (1/cube root of x)] ?
How to find it ?

Answer 1

$$\begin{gathered} \mathrm{Consider}\mathrm{the}\mathrm{following}\mathrm{expression}. \\ \sqrt[3]{3\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)}=2 \\ \mathrm{Note}\mathrm{that}\sqrt[3]{\mathrm{x}}={\mathrm{x}}^{\frac{1}{3}}\mathrm{which}\mathrm{gives}, \\ {\left\{3\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)\right\}}^{\frac{1}{3}}=2 \\ \mathrm{Take}\mathrm{cube}\mathrm{on}\mathrm{both}\mathrm{sides}\mathrm{to}\mathrm{get}, \\ {\left[{\left\{3\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)\right\}}^{\frac{1}{3}}\right]}^3={\left(2\right)}^3 \\ {\left\{3\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)\right\}}^{\frac{1}{3}\times 3}=8 \\ {\left\{3\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)\right\}}^1=8 \\ 3\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)=8 \\ \sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}=\frac{8}{3} \\ \mathrm{Again}\mathrm{cube}\mathrm{both}\mathrm{sides}\mathrm{to}\mathrm{get}, \\ {\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)}^3={\left(\frac{8}{3}\right)}^3 \\ \mathrm{Use}\mathrm{the}\mathrm{algebraic}\mathrm{formula}. \\ {\left(\mathrm{a}-\mathrm{b}\right)}^3={\mathrm{a}}^3-{\mathrm{b}}^3-3\mathrm{ab}\left(\mathrm{a}-\mathrm{b}\right) \\ \mathrm{This}\mathrm{implies}\mathrm{that}, \\ {\left(\sqrt[3]{\mathrm{x}}\right)}^3-{\left(\frac{1}{\sqrt[3]{\mathrm{x}}}\right)}^3-3\left(\sqrt[3]{\mathrm{x}}\right)\left(\frac{1}{\sqrt[3]{\mathrm{x}}}\right)\left(\sqrt[3]{\mathrm{x}}-\frac{1}{\sqrt[3]{\mathrm{x}}}\right)=\frac{{\left(8\right)}^3}{{\left(3\right)}^3} \\ \mathrm{x}-\frac{1}{\mathrm{x}}-3\left(1\right)\left(\frac{8}{3}\right)=\frac{512}{27} \\ \mathrm{x}-\frac{1}{\mathrm{x}}-8=\frac{512}{27} \\ \mathrm{x}-\frac{1}{\mathrm{x}}=\frac{512}{27}+8 \\ \frac{{\mathrm{x}}^2-1}{\mathrm{x}}=\frac{728}{27} \\ \mathrm{Cross}\mathrm{multiply}\mathrm{and}\mathrm{simplify}\mathrm{further}\mathrm{to}\mathrm{get}, \\ 27{\mathrm{x}}^2-27=728\mathrm{x} \\ 27{\mathrm{x}}^2-729\mathrm{x}-27=0 \\ 27{\mathrm{x}}^2-729\mathrm{x}+\mathrm{x}-27=0 \\ 27\mathrm{x}\left(\mathrm{x}-27\right)+1\left(\mathrm{x}-27\right)=0 \\ \left(27\mathrm{x}+1\right)\left(\mathrm{x}-27\right)=0 \\ \mathrm{This}\mathrm{implies}\mathrm{that}, \\ 27\mathrm{x}+1=0 \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{27}}\mathbf{} \\ \mathrm{or} \\ \mathbf{}\mathbf{}\mathrm{x}-27=0 \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{27} \end{gathered}$$