MCQ
If $\text{x}+\frac{1}{\text{x}}=3,$ then $\text{x}^6+\frac{1}{\text{x}^6}=$
- A$927$
- B$414$
- C$364$
- ✓$322$
✓
Answer
Correct option: D.
$322$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^2=\text{x}^2+\frac{1}{\text{x}^2}+2$
$\text{x}+\frac{1}{\text{x}}=3$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(3)^2-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7\ ...(1)$
Cubing both side of equation $(1)$. we have
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^3=(7)^3$
$\Rightarrow(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3(\text{x}^2)\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}+3(7)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=343-21$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=322$
Hence, correct option is $(d)$.
$\text{x}+\frac{1}{\text{x}}=3$ (given)
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=(3)^2-2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=7\ ...(1)$
Cubing both side of equation $(1)$. we have
$\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)^3=(7)^3$
$\Rightarrow(\text{x}^2)^3+\Big(\frac{1}{\text{x}^2}\Big)^3+3(\text{x}^2)\frac{1}{\text{x}^2}\Big(\text{x}^2+\frac{1}{\text{x}^2}\Big)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}+3(7)=7^3$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=343-21$
$\Rightarrow\text{x}^6+\frac{1}{\text{x}^6}=322$
Hence, correct option is $(d)$.
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