Question

if x^3 - 1/x^3 = 14 find x - 1/x

Answer 1

Answer:

⇒ $x - \frac{1}{x} = \sqrt[3]{(7 + 5\sqrt{2})} - \sqrt[3]{(5\sqrt{2} - 7)}$

Step-by-step explanation:

Given :

$x^3-\frac{1}{x^3}=14$

Find the value of :

$x-\frac{1}{x}$

Solution :

⇒ $x^3-\frac{1}{x^3}=14$

⇒ $\frac{x^6-1}{x^3}=14$

⇒ $x^6-1=14x^3$

⇒ $x^6 - 14x^3 - 1 = 0$

As,

⇒ $[-(7 - 5\sqrt{2})] + [-(7 + 5\sqrt{2})] =-14$,

_

⇒ $[-(7 - 5\sqrt{2})]\times[-(7 + 5\sqrt{2})]$

⇒ $[-7 + 5\sqrt{2}]\times[-7 - 5\sqrt{2}]$

⇒ $(-7)^2 - (5\sqrt{2})^2$

⇒ $(-7)^2 - (5)^2(\sqrt{2})^2$

⇒ $49 - 25(2) = 49 - 50 = -1$

So,

⇒ $x^6 - (7 + 5\sqrt{2})x^3 -(7 - 5\sqrt{2}) x^3 - 1 = 0$

⇒ $x^3[x^3 - (7 + 5\sqrt{2})] -(7 - 5\sqrt{2})[x^3 - (7+5\sqrt{2})] = 0$

⇒ $[x^3 - (7 + 5\sqrt{2})][x^3 -(7 - 5\sqrt{2})] = 0$

⇒ $x^3 - (7 + 5\sqrt{2}) = 0$ (or) $x^3 - (7 - 5\sqrt{2}) = 0$

⇒  $x^3 = (7 + 5\sqrt{2})$ (or) $x^3 = (7 - 5\sqrt{2})$

⇒ $x = \sqrt[3]{(7 + 5\sqrt{2})}$ (or) $x = \sqrt[3]{(7 - 5\sqrt{2})}$

⇒ $x = \sqrt[3]{(7 + 5\sqrt{2})}$

⇒ $\frac{1}{x} = \frac{1}{\sqrt[3]{(7 + 5\sqrt{2})}} = \frac{1}{\sqrt[3]{(7 + 5\sqrt{2})}} \times \frac{\sqrt[3]{(7 - 5\sqrt{2})}}{\sqrt[3]{(7 - 5\sqrt{2})}} = \frac{\sqrt[3]{(7 - 5\sqrt{2})}}{-1}=-\sqrt[3]{(7 - 5\sqrt{2})}$

⇒ $x - \frac{1}{x} = \sqrt[3]{(7 + 5\sqrt{2})} - \sqrt[3]{-(7 - 5\sqrt{2})}$

⇒ $x - \frac{1}{x} = \sqrt[3]{(7 + 5\sqrt{2})} - \sqrt[3]{(5\sqrt{2} - 7)}$