Question : If $x+\frac{1}{x}=\sqrt{3}$, the value of $\left (x^{3}+\frac{1}{x^{3}} \right )$ is:
Option 1: $\sqrt{3}$
Option 2: $\frac{1}{\sqrt{3}}$
Option 3: $0$
Option 4: $1$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $0$
Solution : Given: $x+\frac{1}{x}=\sqrt{3}$ Cubing both sides and simplifying the equation, ⇒ $(x+\frac{1}{x})^3=(\sqrt{3})^3$ ⇒ $x^3+\frac{1}{x^3}+3(x×\frac{1}{x})(x+\frac{1}{x})=(\sqrt{3})^3$ ⇒ $x^3+\frac{1}{x^3}+3(x+\frac{1}{x})=3\sqrt{3}$ Putting the value of $x+\frac{1}{x}=\sqrt{3}$ ⇒ $x^3+\frac{1}{x^3}+3\sqrt{3}=3\sqrt{3}$ ⇒ $x^3+\frac{1}{x^3}=3\sqrt{3}-3\sqrt{3}$ ⇒ $x^3+\frac{1}{x^3}=0$ Hence, the correct answer is $0$.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : If $x+\left [\frac{1}{(x+7)}\right]=0$, what is the value of $x-\left [\frac{1}{(x+7)}\right]$?
Question : If $\left(x-\frac{1}{x}\right)^2=12$, what is the value of $\left(x^2-\frac{1}{x^2}\right)$, given that $x>0$?
Question : If $\left(x^2+\frac{1}{x^2}\right)=7$, and $0<x<1$, find the value of $x^2-\frac{1}{x^2}$.
Question : If $\left(x^2 - \frac{1}{x^2}\right) = 4 \sqrt{6}$ and $x>1$, what is the value of $\left(x^3 - \frac{1}{x^3}\right)?$
Question : If $\left(x+\frac{1}{x}\right)=2 \sqrt{2}$ and $x>1$, what is the value of $\left(x^6-\frac{1}{x^6}\right)$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile