Question

if x is = 3 + 2 under root 2 then find the value of (x-1/x)power 3​

Answer 1

Question :

★ If x = 3+2√2 then find the value of

• (x - ¹/ₓ)³

Solution :

$\sf \pink{x=3+2\sqrt{2}}$

$\to \sf \dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}$

Let's rationalize the denominator ,

$\to \sf \dfrac{1}{x}=\dfrac{1}{3+2\sqrt{2}}\times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}}$

• ( a + b ) ( a - b ) = a² - b²

$\to \sf \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{(3)^2-(2\sqrt{2})^2}\\\\\to \sf \dfrac{1}{x}=\dfrac{3-2\sqrt{2}}{9-8}\qquad \; \\\\\to \sf \green{\dfrac{1}{x}=3-2\sqrt{2}}\ \qquad$

Let's solve for our required ,

$\to \sf \bigg(x-\dfrac{1}{x}\bigg)^3$

$\to \sf ((3+2\sqrt{2})-(3-2\sqrt{2}))^3\\\\\to \sf (3+2\sqrt{2}-3+2\sqrt{2})^3\ \; \; \; \; \;$

$\to \sf (4\sqrt{2})^3$

$\leadsto \sf 128\sqrt{2}\ \; \orange{\bigstar}$

Answer 2

$\sf{ x = 3 + 2\sqrt{2} \quad \dots(1)}$

$\implies \sf{\dfrac{1}{x} = \dfrac{1}{3 + 2\sqrt{2}}}$

On rationalising,

$\implies \sf{\dfrac{1}{x} = \dfrac{1}{3 + 2\sqrt{2}} \times \dfrac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}}}$

$\green{\huge{\star}} \boxed{\sf{(a - b)(a + b) = a^2 - b^2}}$

$\implies \sf{\dfrac{1}{x} = \dfrac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2}}$

$\implies \sf{\dfrac{1}{x} = \dfrac{3 - 2\sqrt{2}}{9 - 8}}$

$\implies \sf{\dfrac{1}{x} = 3 - 2\sqrt{2}\quad\dots(2)}$

$\underline{\mathfrak{\red{ (1) - (2) }:-}}$

$\sf{x - \dfrac{1}{x} = 3 + 2\sqrt{2} - (3 - 2\sqrt{2})}$

$\implies \sf{x - \dfrac{1}{x} = 4\sqrt{2}}$

Cubing both sides,

$\sf{ (x - \dfrac{1}{x})^3 = (4\sqrt{2})^3}$

$\implies \sf{ (x - \dfrac{1}{x})^3 = 4^3 \times (\sqrt{2})^3}$

$\leadsto \underline{\boxed{\mathfrak{ \pink{(x - \dfrac{1}{x})^3 = 128\sqrt{2}}}}}$