Question

Find x from the eq.
x=[cube root(a+1) + cube root(a-1)]/[cube root(a+1) - cube root(a-1)

Answer 1

$$\begin{gathered} x=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}}{\sqrt[3]{a+1}-\sqrt[3]{a-1}} \\ \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}+\sqrt[3]{a-1}+\sqrt[3]{a+1}-\sqrt[3]{a-1}}{\sqrt[3]{a+1}+\sqrt[3]{a-1}-\left(\sqrt[3]{a+1}-\sqrt[3]{a-1}\right)}(\mathrm{Applying}\mathrm{componendo}\mathrm{and}\mathrm{dividendo}) \\ \Rightarrow \frac{x+1}{x-1}=\frac{\sqrt[3]{a+1}}{\sqrt[3]{a-1}} \\ \mathrm{Cubing}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \frac{{\left(x+1\right)}^3}{{\left(x-1\right)}^3}=\frac{a+1}{a-1} \\ \mathrm{Again}\mathrm{applying}\mathrm{componendo}\mathrm{and}\mathrm{dividendo},\mathrm{we}\mathrm{get} \\ \frac{{\left(x+1\right)}^3+{\left(x-1\right)}^3}{{\left(x+1\right)}^3-{\left(x-1\right)}^3}=\frac{a+1+a-1}{a+1-\left(a-1\right)} \\ \Rightarrow \frac{2x^3+6x}{6x^2+2}=a \\ \Rightarrow \frac{x^3+3x}{3x^2+1}=a \\ \Rightarrow x^3\mathit{-}3a{x}^{\mathit{2}}+3x\mathit{-}a=0 \end{gathered}$$