Question

If `x - 1/x = 3 + 2sqrt2`, find the value of `x^3 - 1/x^3`

Answer 1

In the given problem, we have to find the value of  `x^3 - 1/x^3`

Given  `x-1/x = 3 + 2sqrt2`

Cubing on both sides of  `x-1/x = 3 + 2sqrt2`

we get \[\left( x - \frac{1}{x} \right)^3 = \left( 3 + 2\sqrt{2} \right)^3\]

We shall use identity `(a+b)^3 = a^3+b^3 + 3ab (a+b)`

`(3+ 2sqrt2)^3 = x^3 -1/x^3- 3 xx x xx 1/x(x- 1/x)`

`3^3 + (2 sqrt2)^3 +3 xx3 xx 2sqrt2(3+2sqrt2) = x^3 - 1/x^3 - 3xx x xx 1/x xx (3+2sqrt2)`

                `27 + 16sqrt2 + 18 sqrt2(3+2sqrt2) = x^3 - 1/x^3 - 3(3+2sqrt2)`

  \[27 + 16\sqrt{2} + 18\sqrt{2} \times 3 + 18\sqrt{2} \times 2\sqrt{2} = x^3 - \frac{1}{x^3} - 9 - 6\sqrt{2}\]

                              \[27 + 16\sqrt{2} + 54\sqrt{2} + 72 = x^3 - \frac{1}{x^3} - 9 - 6\sqrt{2}\]

     `27+ 16sqrt2 + 54sqrt2 + 72 + 9+ 6sqrt2 = x^3 - 1/x^3`

`[27 + 72 + 9]+[16sqrt2 + 54 sqrt2 +6sqrt2] = x^3 - 1/x^3`

                                                `108 + 76 sqrt2 = x^3 - 1/x^3`

Hence the value of  `x^3-1/x^3`is  `108+76sqrt2`.

Answer 2

`(x - 1/x)^3 = (x^3 - 1/x^3) =- 3 · x · 1/x · (x - 1/x)`

`x^3 - 1/x^3 = (x - 1/x)^3 + 3(x - 1/x)`

`x^3 - 1/x^3 = (3 + 2sqrt2)^3 + 3(3 + 2sqrt2)`

= `3 + 2sqrt2 ((3 + 2sqrt2)^2 + 3)`

= `(3 + 2sqrt2) (9 + 8 + 12sqrt2 + 3)`

= `63 + 36sqrt2 + 42sqrt2 + 24 · 2`

= `63 + 48 + 78sqrt2`

= `111 + 78sqrt2`.