CAT 2025 Slot 3 QA Question 2: If $(x^{2} + \dfrac{1}{x^{2}}) = 25$ and $x 0$, then the value of $(x^{7} + \dfrac{1}{x^{7}})$ is

Question

If $(x^{2} + \dfrac{1}{x^{2}}) = 25$ and $x > 0$, then the value of $(x^{7} + \dfrac{1}{x^{7}})$ is

Accepted Answer

Given $(x^{2} + \dfrac{1}{x^{2}}) = 25$. Since $x$ is positive, we can find $x + \dfrac{1}{x}$.
$(x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2$
$(x + \dfrac{1}{x})^{2} = 25 + 2 = 27$
$x + \dfrac{1}{x} = 3 	imes \sqrt{3}$
Let $a_n = x^n + \dfrac{1}{x^n}$. Then $a_{n+1} = a_1a_n - a_{n-1}$.
$a_1 =

If (x2+1x2)=25(x^{2} + \dfrac{1}{x^{2}}) = 25(x2+x21​)=25 and x>0x > 0x>0, then the value of (x7+1x7)(x^{7} + \dfrac{1}{x^{7}})(x7+x71​) is

If $(x^{2} + \dfrac{1}{x^{2}}) = 25$ and $x > 0$ , then the value of $(x^{7} + \dfrac{1}{x^{7}})$ is