Question

If \(x\) is a positive real number such that \(x^8+\bigg(\frac{1}{x}\bigg)^8=47\) , then the value of \(x^9+\bigg(\frac{1}{x}\bigg)^9\) is

• A. \(34\sqrt 5\)
• B. \(40\sqrt5\)
• C. \(36\sqrt5\)
• D. \(30\sqrt5\)

Answer 1

The Correct Option is A

Given:   $x^8 + \left(\frac{1}{x}\right)^8 = 47$

We can write this as: 

$\Rightarrow (x^4)^2 + \left(\frac{1}{x^4}\right)^2 = 47$
$\Rightarrow \left(x^4 + \frac{1}{x^4}\right)^2 - 2 = 47$
$\Rightarrow \left(x^4 + \frac{1}{x^4}\right)^2 = 49$
$\Rightarrow x^4 + \frac{1}{x^4} = 7$    …… (i)

Now, from equation (i):

$\Rightarrow (x^2)^2 + \left(\frac{1}{x^2}\right)^2 = 7$
$\Rightarrow \left(x^2 + \frac{1}{x^2}\right)^2 - 2 = 7$
$\Rightarrow \left(x^2 + \frac{1}{x^2}\right)^2 = 9$
$\Rightarrow x^2 + \frac{1}{x^2} = 3$    …… (ii)

Now assume: $x + \frac{1}{x} = \sqrt{5}$

Then:

$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right)$
$= (\sqrt{5})^3 - 3\sqrt{5} = 5\sqrt{5} - 3\sqrt{5} = 2\sqrt{5}$  …… (iii)

Now calculate: $x^9 + \frac{1}{x^9}$

$x^9 + \frac{1}{x^9} = \left(x^3 + \frac{1}{x^3}\right)^3 - 3\left(x^3 + \frac{1}{x^3}\right)$
$= (2\sqrt{5})^3 - 3(2\sqrt{5})$
$= 8 \cdot 5\sqrt{5} - 6\sqrt{5}$
$= 40\sqrt{5} - 6\sqrt{5}$
$= 34\sqrt{5}$

Therefore, the correct answer is:   (A) $34\sqrt{5}$