Question
If $\text{x}^3+\frac{1}{\text{x}^3}=110,$ then $\text{x}+\frac{1}{\text{x}}=$
✓
Answer
- 5
Solution:
$\Big(\text{x}+\frac{1}{\text{x}}\Big)^3=\text{x}^3+\frac{1}{\text{x}^3}+3\Big(\text{x}+\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=\text{x}^3+\frac{1}{\text{x}^3}$
$\Rightarrow\Big(\text{x}+\frac{1}{\text{x}}\Big)^3-3\Big(\text{x}+\frac{1}{\text{x}}\Big)=110$
Let $\text{x}+\frac{1}{\text{x}}=\text{t}$
$\Rightarrow\text{t}^3-3\text{t}-110=0$
t =5 is one of it's solution which is real, other two solutions are imaginary
$\Rightarrow\text{x}+\frac{1}{\text{x}}=5$
Hence, correct option is (a).
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