MCQ
If $\text{x}^3-\frac{1}{\text{x}^3}=14,$ then $\text{x}-\frac{1}{\text{x}}=$
- A$5$
- B$4$
- C$3$
- ✓$2$
✓
Answer
Correct option: D.
$2$
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow \mathrm{t}^3+3 \mathrm{t}-14=0$
$\Rightarrow \mathrm{t}^3-2 \mathrm{t}^2+2 \mathrm{t}^2-4 \mathrm{t}+7 \mathrm{t}-14=0$
$\Rightarrow \mathrm{t}(\mathrm{t}-2)+2 \mathrm{t}(\mathrm{t}-2)+7(\mathrm{t}-2)=0$
$\Rightarrow(\mathrm{t}-2)(\mathrm{t}+2 \mathrm{t}+7)=0$
$\Rightarrow \mathrm{t}^2+2 \mathrm{t}+7=0 \text { has no real roots }$
So, $t=2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
Hence, correct option is $(d)$.
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