Question Number 14658 by tawa tawa last updated on 03/Jun/17
$$\mathrm{Given}\:\mathrm{that}: \\ $$$$\mathrm{x}\:=\:\left(\sqrt{\mathrm{2}}\:+\:\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \:−\:\left(\sqrt{\mathrm{2}}\:−\:\mathrm{1}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\mathrm{Show}\:\mathrm{that}\:,\:\:\:\:\:\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{3x}\:=\:\mathrm{2} \\ $$
Answered by Tinkutara last updated on 03/Jun/17
![x^3 = (√2) + 1 − (√2) + 1 − 3(1)(x) [Using (a − b)^3 = a^3 − b^3 − 3ab(a − b)] x^3 + 3x = 2](https://www.tinkutara.com/question/Q14659.png)
$${x}^{\mathrm{3}} \:=\:\sqrt{\mathrm{2}}\:+\:\mathrm{1}\:−\:\sqrt{\mathrm{2}}\:+\:\mathrm{1}\:−\:\mathrm{3}\left(\mathrm{1}\right)\left({x}\right) \\ $$$$\left[\mathrm{Using}\:\left({a}\:−\:{b}\right)^{\mathrm{3}} \:=\:{a}^{\mathrm{3}} \:−\:{b}^{\mathrm{3}} \:−\:\mathrm{3}{ab}\left({a}\:−\:{b}\right)\right] \\ $$$${x}^{\mathrm{3}} \:+\:\mathrm{3}{x}\:=\:\mathrm{2} \\ $$
Commented by tawa tawa last updated on 03/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$