MCQ
If $\text{x}+\frac{1}{\text{x}}=2,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$
- A$8$
- B$14$
- C$64$
- ✓$2$
✓
Answer
Correct option: D.
$2$
On cubing we get
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow2^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times2$
$\Rightarrow\text{3}^3+\Big(\frac{1}{\text{x}^3}\Big)=8-6$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=2$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\text{x}\times\frac{1}{\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow2^3=\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)+3\times2$
$\Rightarrow\text{3}^3+\Big(\frac{1}{\text{x}^3}\Big)=8-6$
$\Rightarrow\text{x}^3+\Big(\frac{1}{\text{x}^3}\Big)=2$
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