Question

If $$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$$, prove that $$x = 11$$.

[Hint : By componendo and dividendo, $$\frac{(x + 1)^3}{(x - 1)^3} = \frac{432}{250} = \frac{216}{125} = \left(\frac{6}{5}\right)^3 \Rightarrow \frac{x + 1}{x - 1} = \frac{6}{5}$$]

Answer 1

Given: $$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$$

To prove: $$x = 11$$

Proof:

1. $$\frac{x^3 + 3x}{3x^2 + 1} = \frac{341}{91}$$ [Given]
2. $$\frac{(x^3 + 3x) + (3x^2 + 1)}{(x^3 + 3x) - (3x^2 + 1)} = \frac{341 + 91}{341 - 91}$$ [By Componendo & Dividendo]
3. or, $$\frac{x^3 + 3x^2 + 3x + 1}{x^3 - 3x^2 + 3x - 1} = \frac{432}{250}$$
4. or, $$\frac{(x + 1)^3}{(x - 1)^3} = \frac{216}{125}$$
5. or, $$\left(\frac{x + 1}{x - 1}\right)^3 = \left(\frac{6}{5}\right)^3$$
6. or, $$\frac{x + 1}{x - 1} = \frac{6}{5}$$ [Taking cube root on both sides]
7. or, $$\frac{(x + 1) + (x - 1)}{(x + 1) - (x - 1)} = \frac{6 + 5}{6 - 5}$$ [By Componendo & Dividendo]
8. or, $$\frac{2x}{2} = \frac{11}{1}$$
9. or, $$x = 11$$

Hence proved.