Question

If $$x = \frac{\sqrt[3]{m+1} + \sqrt[3]{m-1}}{\sqrt[3]{m+1} - \sqrt[3]{m-1}}$$, prove that $$x^3 - 3x^2m + 3x - m = 0$$.

\[ \begin{array}{l} [\textbf{Hint :}\ \text{By componendo and dividendo, we have}\ \dfrac{x + 1}{x - 1} = \dfrac{\sqrt[3]{m + 1}}{\sqrt[3]{m - 1}} \Leftrightarrow \dfrac{(x + 1)^{3}}{(x - 1)^{3}} = \dfrac{m + 1}{m - 1}. \\[16pt] \text{Now, apply componendo and dividendo.}] \end{array} \]

Answer 1

Given: $$x = \frac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} - \sqrt[3]{m - 1}}$$

To prove: $$x^3 - 3x^2m + 3x - m = 0$$

Proof:

1. $$\frac{x}{1} = \frac{\sqrt[3]{m + 1} + \sqrt[3]{m - 1}}{\sqrt[3]{m + 1} - \sqrt[3]{m - 1}}$$
2. or, $$\frac{x + 1}{x - 1} = \frac{(\sqrt[3]{m + 1} + \sqrt[3]{m - 1}) + (\sqrt[3]{m + 1} - \sqrt[3]{m - 1})}{(\sqrt[3]{m + 1} + \sqrt[3]{m - 1}) - (\sqrt[3]{m + 1} - \sqrt[3]{m - 1})}$$ [By componendo and dividendo]
3. or, $$\frac{x + 1}{x - 1} = \frac{2\sqrt[3]{m + 1}}{2\sqrt[3]{m - 1}}$$
4. or, $$\frac{x + 1}{x - 1} = \frac{\sqrt[3]{m + 1}}{\sqrt[3]{m - 1}}$$
5. or, $$\frac{(x + 1)^3}{(x - 1)^3} = \frac{m + 1}{m - 1}$$ [On cubing both sides]
6. or, $$\frac{(x + 1)^3 + (x - 1)^3}{(x + 1)^3 - (x - 1)^3} = \frac{(m + 1) + (m - 1)}{(m + 1) - (m - 1)}$$ [By componendo and dividendo]
7. or, $$\frac{(x^3 + 3x^2 + 3x + 1) + (x^3 - 3x^2 + 3x - 1)}{(x^3 + 3x^2 + 3x + 1) - (x^3 - 3x^2 + 3x - 1)} = \frac{2m}{2}$$
8. or, $$\frac{2(x^3 + 3x)}{2(3x^2 + 1)} = m$$
9. or, $$\frac{x^3 + 3x}{3x^2 + 1} = m$$
10. or, $$x^3 + 3x = m(3x^2 + 1)$$
11. or, $$x^3 + 3x = 3x^2m + m$$
12. or, $$x^3 - 3x^2m + 3x - m = 0$$

Hence proved.