Question
If $\text{x}^3-\frac{1}{\text{x}^3}=14,$ then $\text{x}-\frac{1}{\text{x}}=$
✓
Answer
$\Big(\text{x}-\frac{1}{\text{x}}\Big)^3=\text{x}^3-\frac{1}{\text{x}^3}-3\not\text{x}\frac{1}{\not\text{x}}\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow t^3 + 3t - 14 = 0$
$\Rightarrow t^3 - 2t^2 + 2t^2 - 4t + 7t -14 = 0$
$\Rightarrow t(t - 2) + 2t(t - 2) + 7(t - 2) = 0$
$\Rightarrow (t - 2)(t + 2t + 7) = 0$
$\Rightarrow t^2 + 2t + 7 = 0$ has no real roots
So$, t = 2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
$\text{x}^3-\frac{1}{\text{x}^3}=\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-\text{x}^3-\frac{1}{\text{x}^3}=0$
$\Rightarrow\Big(\text{x}-\frac{1}{\text{x}}\Big)^3+3\Big(\text{x}-\frac{1}{\text{x}}\Big)-14=0$
Let $\Rightarrow\text{x}-\frac{1}{\text{x}}=\text{t}$
$\Rightarrow t^3 + 3t - 14 = 0$
$\Rightarrow t^3 - 2t^2 + 2t^2 - 4t + 7t -14 = 0$
$\Rightarrow t(t - 2) + 2t(t - 2) + 7(t - 2) = 0$
$\Rightarrow (t - 2)(t + 2t + 7) = 0$
$\Rightarrow t^2 + 2t + 7 = 0$ has no real roots
So$, t = 2$ is a solution
$\Rightarrow\text{x}-\frac{1}{\text{x}}=2$
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