Question : If $x-\frac{3}{x}=6, x \neq 0$, then the value of $\frac{x^4-\frac{27}{x^2}}{x^2-3 x-3}$ is:
Option 1: 80
Option 2: 270
Option 3: 54
Option 4: 90
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: 90
Solution : $x-\frac{3}{x}=6$ ⇒ $x^2 - 3 = 6x$ ⇒ $x^2 - 6x -3 =0$ ⇒ $x^2 - 3x -3 =3x$ ---(1) ($x-\frac{3}{x})^3= 6^3$ ⇒ $x^3 - (\frac{3}{x})^3 - 3(x-\frac{3}{x}) = 216$ ⇒ $x^3 - (\frac{3}{x})^3 - 9(6) = 216$ ⇒ $x^3 - (\frac{3}{x})^3 - 54 = 216$ ⇒ $x^3 - (\frac{3}{x})^3 = 270$ --- (2) Now, $\frac{x^4-\frac{27}{x^2}}{x^2-3 x-3}$ = $\frac{x^4-\frac{27}{x^2}}{3x}$ [from equation (1)] = $\frac{x^3-\frac{27}{x^3}}{3}$ = $\frac{270}{3}$ [From the above equation (2)] = 90 Hence the correct answer is 90.
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : If $x-\frac{1}{x}=5, x \neq 0$, then what is the value of $\frac{x^6+3 x^3-1}{x^6-8 x^3-1} ?$
Question : If $(x+\frac{1}{x})\neq 0$ and $(x^3+\frac{1}{x^3})= 0$, then the value $(x+\frac{1}{x})^4$ is:
Question : If $(x+\frac{1}{x})^2=3$, then the value of $x^{72}+x^{66}+x^{54}+x^{24}+x^6+1$ is:
Question : If $2x-\frac{2}{x}=1(x \neq 0)$, then the the value of $(x^3-\frac{1}{x^3})$ is:
Question : If $x+\frac{1}{x}=3, x \neq 0$, then the value of $x^7+\frac{1}{x^7}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile