Question : If $x+\frac{1}{x}=2 \sqrt{5}$ where $x>1$, then the value of $x^3-\frac{1}{x^3}$ is:
Option 1: 82
Option 2: 76
Option 3: 86
Option 4: 78
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: 76
Solution : Given: $x+\frac{1}{x}=2 \sqrt{5}$ ⇒ $(x+\frac{1}{x})^2=(2 \sqrt{5})^2$ ⇒ $x^2+\frac{1}{x^2}+2=20$ ⇒ $x^2+\frac{1}{x^2}=18$ -------------(i) ⇒ $x^2+\frac{1}{x^2}-2=18-2=16$ ⇒ $(x-\frac{1}{x})^2=(4)^2$ ⇒ $x-\frac{1}{x}=4$ -------------(ii) So, $x^3-\frac{1}{x^3}$ = $(x-\frac{1}{x})(x^2+\frac{1}{x^2}+x×\frac{1}{x})$ = 4 × (18 + 1) = 76 Hence, the correct answer is 76.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Question : If $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$, what is the value of $x$?
Question : If $x^{2} + \frac{1}{x^{2}} = 18$ and $x > 0$, what is the value of $x^{3} + \frac{1}{x^{3}}?$
Question : If $x=\frac{1}{x-3},(x>0)$, then the value of $x+\frac{1}{x}$ is:
Question : If $x>0$ and $x^4+\frac{1}{x^4}=254$, what is the value of $x^5+\frac{1}{x^5}?$
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile