MCQ
If $\text{x}+\frac{1}{\text{x}}=2,$ then $\text{x}^3+\frac{1}{\text{x}^3}=$
- A$64$
- B$14$
- C$8$
- ✓$2$
✓
Answer
Correct option: D.
$2$
By using identity,
$(a+b)^3=a^3+b^3+3 a b(a+b)$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(\not\text{x})\frac{1}{\not\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)\\=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Now $\text{x}+\frac{1}{\text{x}}=2$
$\Rightarrow(2)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(2)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=(2)^3-3\times2=8-6=2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is $(d)$.
$(a+b)^3=a^3+b^3+3 a b(a+b)$
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(\not\text{x})\frac{1}{\not\text{x}}\Big(\text{x}+\frac{1}{\text{x}}\Big)\\=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
Now $\text{x}+\frac{1}{\text{x}}=2$
$\Rightarrow(2)^3=\text{x}^3+\frac{1}{\text{x}^3}+3(2)$
$\Rightarrow\text{x}^3+\frac{1}{\text{x}^3}=(2)^3-3\times2=8-6=2$
$\Rightarrow\text{x}^2+\frac{1}{\text{x}^2}=23$
Hence, correct option is $(d)$.
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