4 Views

Question : If $x^{2} + \frac{1}{x^{2}} = 18$ and $x > 0$, what is the value of $x^{3} + \frac{1}{x^{3}}?$

Option 1: $36 \sqrt{5}$

Option 2: $40 \sqrt{5}$

Option 3: $46 \sqrt{5}$

Option 4: $34 \sqrt{5}$


Team Careers360 7th Jan, 2024
Answer (1)
Team Careers360 23rd Jan, 2024

Correct Answer: $34 \sqrt{5}$


Solution : $x^{2} + \frac{1}{x^{2}} = 18$
⇒ $x^{2} + \frac{1}{x^{2}} + 2 = 18+2$
⇒ $(x + \frac{1}{x})^{2} = 20$
⇒ $(x + \frac{1}{x}) = \sqrt{20}$
Now, $(x + \frac{1}{x})^{3}=x^3+\frac{1}{x^3}+3 × x × \frac{1}{x} × (x + \frac{1}{x})$
⇒ $(\sqrt{20})^3=x^3+\frac{1}{x^3}+3×\sqrt{20}$
⇒ $x^3+\frac{1}{x^3}=20\sqrt{20} - 3\sqrt{20}$
⇒ $x^3+\frac{1}{x^3}=17\sqrt{20}$
$\therefore x^3+\frac{1}{x^3}=34\sqrt{5}$
Hence, the correct answer is $34\sqrt{5}$.

SSC CGL Complete Guide

Candidates can download this ebook to know all about SSC CGL.

Download EBook

Know More About

Related Questions

TOEFL ® Registrations 2024
Apply
Accepted by more than 11,000 universities in over 150 countries worldwide
Manipal Online M.Com Admissions
Apply
Apply for Online M.Com from Manipal University
GRE ® Registrations 2024
Apply
Apply for GRE® Test now & save 10% with ApplyShop Gift Card | World's most used Admission Test for Graduate & Professional Schools
View All Application Forms

Download the Careers360 App on your Android phone

Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile

150M+ Students
30,000+ Colleges
500+ Exams
1500+ E-books