Question : If $x^2+6x+1=0$, then the value of $(x+6)^3+\frac{1}{(x+6)^3}$ = ?
Option 1: 245
Option 2: 216
Option 3: 186
Option 4: 198
Correct Answer: 198
Solution : $x^2+6x+1=0$ On dividing both sides by $x$, $⇒x+6+\frac{1}{x}=0$ $⇒x+6=-\frac{1}{x}$ $⇒x=\frac{-1}{x+6}$ $⇒x+\frac{1}{x+6}=0$ Adding both sides 6. $⇒x + 6+\frac{1}{x+6}=6$ Cubing both sides of the equation Now, $(x+6)^3+\frac{1}{(x+6)^3} = 6^3-3\times 6$ $=198$ Hence, the correct answer is 198.
Application | Eligibility | Selection Process | Result | Cutoff | Admit Card | Preparation Tips
Question : If $x^4+\frac{1}{x^4}=1154$, where $x>0$, then what is the value of $x^3+\frac{1}{x^3}?$
Option 1: 205
Option 2: 214
Option 3: 185
Question : if $x+\frac{1}{x}=2$, then the value of $x^4+\frac{1}{x^4}$=__________.
Option 1: 0
Option 2: 2
Option 3: –1
Option 4: 1
Question : If $\frac{1}{x+2}=\frac{3}{y+3}=\frac{1331}{z+1331}=\frac{1}{3}$, then what is the value of $\frac{x}{x+1}+\frac{y}{y+6}+\frac{z}{z+2662}$?
Option 1: $0$
Option 2: $1$
Option 3: $\frac{3}{2}$
Option 4: $3$
Question : If $x + \frac{1}{x} = \sqrt{3}$, then the value of $x^{18} + x^{12} + x^{6} + 1$ is:
Option 2: 1
Option 3: 2
Option 4: 3
Question : If $\frac{x^{2}-x+1}{x^{2}+x+1}=\frac{2}{3}$, then the value of $\left (x+\frac{1}{x} \right)$ is:
Option 1: 4
Option 2: 5
Option 3: 6
Option 4: 8
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile