Question

x-1/x =3+2 root 2, find x cube- 1/x cube​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\sf{{x}^{3} - \dfrac{1}{ {x}^{3} } =76 \sqrt{2} +108}}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

Given :- $\sf{x - \dfrac{1}{x} = 3 + 2 \sqrt{2} }$

To find : $\sf{x^3 - \dfrac{1}{x^3} }$

Solution :-

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

By cubing on both sides

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

In LHS - We know that (x - y)³ = x³ - y³ - 3xy(x - y)

Here x = x, y = 1/x

By substituting the values in the identity we have,

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

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

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

[Since $\bf{x - \dfrac{1}{x} = 3 + 2 \sqrt{2} }$ ]

${x}^{3} - \dfrac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} = {(3 + 2 \sqrt{2}) }^{3}$

In RHS - We know that (x + y)³ = x³ + y³ + 3xy(x + y)

Here x = 3, y = 2√2

By substituting the values in the identity we have,

${x}^{3} - \dfrac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} = {3}^{3} + {(2 \sqrt{2})}^{3} + 3(3)(2 \sqrt{2})(3 + 2 \sqrt{2})$

${x}^{3} - \dfrac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} =27 +8(2 \sqrt{2}) + 18 \sqrt{2}(3 + 2 \sqrt{2})$

${x}^{3} - \dfrac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} =27 +16 \sqrt{2} + 54 \sqrt{2} +36(2)$

${x}^{3} - \dfrac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} =27 +70 \sqrt{2} +72$

${x}^{3} - \dfrac{1}{ {x}^{3} } - 9 - 6 \sqrt{2} =70 \sqrt{2} +99$

${x}^{3} - \dfrac{1}{ {x}^{3} } =70 \sqrt{2} +99 + 9 + 6 \sqrt{2}$

${x}^{3} - \dfrac{1}{ {x}^{3} } =76 \sqrt{2} +108$

$\Huge{\boxed{\sf{{x}^{3} - \dfrac{1}{ {x}^{3} } =76 \sqrt{2} +108}}}$

Answer 2

ANSWER:

given

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

REQUIRED TO FIND:

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

METHOD:

in the attachment

we just took the given information as an equation

and then do cubic on both sides

IDENTITIES USED:

${(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)$