MCQ
If $\text{x}+\frac{1}{\text{x}}=3,$ then $\text{x}^6+\frac{1}{\text{x}^6}=$
- A$927$
- ✓$322$
- C$414$
- D$364$
✓
Answer
Correct option: B.
$322$
On cubing we get.
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow27=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times\text{3}$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=27-9$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=18$
Now, $\Big(\text{x}^3+\frac{1}{\text{x}^3}\Big)^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2\times\text{x}^3\times\frac{1}{\text{x}^3}$
$\Rightarrow18^2=\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)+2$
$\text{x}^6+\Big(\frac{1}{\text{x}^6}\Big)=324-2=322$
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