CAT 2023 Slot 3 QA Question 2: If x is a positive real number such that $x^8 + \left(\dfrac{1}{x}\right)^8 = 47$, then the value of $x^9 + \left(\dfrac{1}{x}\right)^9$ is

Question

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

Accepted Answer

We repeatedly use $a^2 + b^2 = (a + b)^2 - 2ab$ with $a = x^k$ and $b = \dfrac{1}{x^k}$.

$x^8 + \dfrac{1}{x^8} = 47$
$\left(x^4 + \dfrac{1}{x^4}\right)^2 - 2 = 47$
$\left(x^4 + \dfrac{1}{x^4}\right)^2 = 49$
$x^4 + \dfrac{1}{x^4} = 7$

$\left(x^2 + \dfrac{1}{x^2}\right)^2 - 2 = 7$
$x^2 + \dfrac{1}{

If x is a positive real number such that x8+(1x)8=47x^8 + \left(\dfrac{1}{x}\right)^8 = 47x8+(x1​)8=47, then the value of x9+(1x)9x^9 + \left(\dfrac{1}{x}\right)^9x9+(x1​)9 is