Question

if x=√3+√2then find xcube-1/xcube equal to
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Answer 1

Question :-

If $\bf{x = \sqrt{3} + \sqrt{2}}$ , then find the value of $\bf{x^{3} - \dfrac{1}{x^{3}}}$

To Find :-

The value of :-

$\bf{x^{3} - \dfrac{1}{x^{3}}}$

Given :-

• $\bf{x = \sqrt{3} + \sqrt{2}}$

We know :-

• $\bf{a^{3} - b^{3} = a^{3} - b^{3} - 3ab(a - b)}$

• If $\bf{x = \sqrt{3} + \sqrt{2}}$ ,then $\bf{\dfrac{1}{x} = \sqrt{3} - \sqrt{2}}$

• $\bf{a^{2} - b^{2} = (a + b)(a - b)}$

Concept :-

To Prove that :-

If $\bf{x = \sqrt{3} + \sqrt{2}}$ ,then $\bf{\dfrac{1}{x} = \sqrt{3} - \sqrt{2}}$

$:\implies \bf{\dfrac{1}{x} = \dfrac{1}{\sqrt{3} + \sqrt{2}}} \\ \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{1}{\sqrt{3} + \sqrt{2}} \times \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}} \\ \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{\sqrt{3} - \sqrt{2}}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}} \\ \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{\sqrt{3} - \sqrt{2}}{\sqrt{3}^{2} - \sqrt{2}^{2}}} \\ \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{\sqrt{3} - \sqrt{2}}{3 - 2}} \\ \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \dfrac{\sqrt{3} - \sqrt{2}}{1}} \\ \\ \\ \\ :\implies \bf{\dfrac{1}{x} = \sqrt{3} - \sqrt{2}} \\ \\ \\ \\ \therefore \purple{\bf{\dfrac{1}{x} = \sqrt{3} - \sqrt{2}}}$

Hence, Proved ✓✓ !!

Now using this information ,we can find the required value.

Solution :-

• Equation (i) = $\bf{x = \sqrt{3} + \sqrt{2}}$

• Equation (ii) = $\bf{\dfrac{1}{x} = \sqrt{3} - \sqrt{2}}$

On subtracting Eq.(ii) from Eq.(i) , we get :-

$:\implies \bf{x - \dfrac{1}{x}} \\ \\ \\ \\ :\implies \bf{\sqrt{3} + \sqrt{2} - (\sqrt{3} - \sqrt{2})} \\ \\ \\ \\ :\implies \bf{\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}} \\ \\ \\ \\ :\implies \bf{\not{\sqrt{3}} + \sqrt{2} - \not{\sqrt{3}} + \sqrt{2}} \\ \\ \\ \\ :\implies \bf{\sqrt{2} + \sqrt{2}} \\ \\ \\ \\ :\implies \bf{2\sqrt{2}} \\ \\ \\ \\ \purple{\bf{x - \dfrac{1}{x} = 2\sqrt{2}}}$

Hence, $\bf{x - \dfrac{1}{x} = 2\sqrt{2}}$

Now , by cubing $\bf{x - \dfrac{1}{x}}$ , we get :-

$:\implies \bf{x - \dfrac{1}{x}} \\ \\ \\ \\ :\implies \bf{\bigg(x - \dfrac{1}{x}\bigg)^{3}} \\ \\ \\ \\ :\implies \bf{\bigg(x - \dfrac{1}{x})^{3} = x^{3} - \bigg(\dfrac{1}{x}\bigg)^{3} - 3 \times x \times \dfrac{1}{x} \times \bigg(x - \dfrac{1}{x}\bigg)} \\ \\ \\ \\ :\implies \bf{\bigg(x - \dfrac{1}{x})^{3} = x^{3} - \bigg(\dfrac{1}{x}\bigg)^{3} - 3 \times \not{x} \times \dfrac{1}{\not{x}} \times \bigg(x - \dfrac{1}{x}\bigg)} \\ \\ \\ \\ :\implies \bf{\bigg(x - \dfrac{1}{x}\bigg)^{3} = x^{3} - \bigg(\dfrac{1}{x}\bigg)^{3} - 3 \times \bigg(x - \dfrac{1}{x}\bigg)}$

On putting the value of $\bf{x - \dfrac{1}{x}}$ , in the equation , we get :-

$:\implies \bf{(2\sqrt{2})^{3} = x^{3} - \bigg(\dfrac{1}{x}\bigg)^{3} - 3 \times (2\sqrt{2})} \\ \\ \\ \\ :\implies \bf{16\sqrt{2} = x^{3} - \bigg(\dfrac{1}{x}\bigg)^{3} - 6\sqrt{2}} \\ \\ \\ \\ :\implies \bf{- \bigg(x^{3} - \dfrac{1}{x^{3}}\bigg) = - 16\sqrt{2} - 6\sqrt{2}} \\ \\ \\ \\ :\implies \bf{- \bigg(x^{3} - \dfrac{1}{x^{3}}\bigg) = - 22\sqrt{2}} \\ \\ \\ \\ :\implies \bf{\not{-} \bigg(x^{3} - \dfrac{1}{x^{3}}\bigg) = \not{-} 22\sqrt{2}} \\ \\ \\ \\ :\implies \bf{x^{3} - \dfrac{1}{x^{3}} = 22\sqrt{2}} \\ \\ \\ \\ :\implies \purple{\bf{x^{3} - \dfrac{1}{x^{3}} = 22\sqrt{2}}}$

Hence, the value of $\bf{x^{3} - \dfrac{1}{x^{3}}}$ is $\bf{22\sqrt{2}}$