Question

this is the question pls answer​

Answer 1

Answer:

Given:

• $x+ \dfrac{1}{x} = 5$
• $x^3 - \dfrac{1}{x^3}$

Solution:

$\implies x + \dfrac{1}{x} = 5$

$\implies \left(x + \dfrac{1}{x} \right)^2 = 5^2$

$\implies x^2 + \dfrac{1}{x^2} + 2= 25$

$\implies x^2 + \dfrac{1}{x^2} = 23$

Finding $\underline{\bf{x - \dfrac{1}{x}}}$:

$\implies x^2 + \dfrac{1}{x^2} - 2 = 23 - 2$

$\implies \left( x - \dfrac{1}{x}\right)^2 = 21$

$\implies x - \dfrac{1}{x} = \sqrt{21}$

Solving further:

$\implies \left( x - \dfrac{1}{x} \right)^3 = (\sqrt{21})^3$

$\implies x^3 - \dfrac{1}{x^3} - 3 \left(x - \dfrac{1}{x}\right) = 21\sqrt{21}$

$\implies x^3 - \dfrac{1}{x^3} -3(\sqrt{21})= 21\sqrt{21}$

$\implies x^3 - \dfrac{1}{x^3} = 21 \sqrt{21} + 3\sqrt{21}$

$\boxed{\implies{\boxed{x^3 - \dfrac{1}{x^3} = 24 \sqrt{21}}}}$