Question : If $x+\frac{1}{x}=\sqrt{3}$ then, the value of $x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$ is:
Option 1: $\sqrt{3}$
Option 2: $-\sqrt{3}$
Option 3: $1$
Option 4: $0$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $0$
Solution : Given: $x+\frac{1}{x}=\sqrt{3}$ Cubing both sides, we get $x^3+\frac{1}{x^3}+3(x)(\frac{1}{x})(x+\frac{1}{x})=3\sqrt{3}$ ⇒ $x^3+\frac{1}{x^3}+3\sqrt{3}=3\sqrt{3}$ ⇒ $x^6+1=0$ The expression is $x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$ $=x^{30}+x^{24}+x^{18}+x^{12}$ $=x^{30}+x^{24}+x^{12}(x^{6}+1)$ $=x^{30}+x^{24}$ $=x^{24}(x^{6}+1)$ $=0$ Hence, the correct answer is $0$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : If $x=(\sqrt{6}-1)^{\frac{1}{3}}$, then the value of $\left(x-\frac{1}{x}\right)^3+3\left(x-\frac{1}{x}\right)$ is:
Question : If $x^{2}+\frac{1}{x^{2}}=1$, then, what is the value of $x^{48}+x^{42}+x^{36}+x^{30}+x^{24}+x^{18}+x^{12}+x^{6}+1$?
Question : If $x=\sqrt{3}-\frac{1}{\sqrt{3}}, y=\sqrt{3}+\frac{1}{\sqrt{3}}$, then the value of $\frac{x^2}{y}+\frac{y^2}{x}$ is:
Question : If $x=\frac{4\sqrt{15}}{\sqrt{5}+\sqrt{3}}$, the value of $\frac{x+\sqrt{20}}{x–\sqrt{20}}+\frac{x+\sqrt{12}}{x–\sqrt{12}}$ is:
Question : If $x=\frac{1}{x-5}(x>0)$, then the value of $x+\frac{1}{x}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile