If-x-1-3-3-x-36-1-3-then-x-2-1-x-

Question Number 173997 by mnjuly1970 last updated on 22/Jul/22

I f

\sqrt[3]{x} - 3 = \sqrt[3]{x - 36}

t h e n

\frac{x^{2} - 1}{x} = ?

Answered by behi834171 last updated on 23/Jul/22

x = t^{3}, x - 36 = s^{3}

\Rightarrow {\begin{cases} t - s = 3 \Rightarrow (t - s) [{(t - s)}^{2} + 3 t s] = 36 \Rightarrow \\ t^{3} - s^{3} = 36 \end{cases}

3 (9 + 3 t s) = 36 \Rightarrow t s = 1 \Rightarrow t + s = \pm {[{(t - s)}^{2} + 4 t s]}^{\frac{1}{2}} =

\Rightarrow t + s = \pm \sqrt{13}

\Rightarrow z^{2} - (\pm \sqrt{13}) z + 1 = 0

\Rightarrow z = \frac{\pm \sqrt{13} \pm 3}{2} \Rightarrow {\begin{cases} s = \frac{\pm \sqrt{13} + 3}{2} \\ t = \frac{\pm \sqrt{13} - 3}{2} \end{cases}

x = t^{3} = {(\frac{\pm \sqrt{13} + 3}{2})}^{3} = {\begin{cases} \frac{13 \sqrt{13} + 117 + 27 \sqrt{13} + 27}{8} \\ \frac{- 13 \sqrt{13} + 117 - 27 \sqrt{13} + 27}{8} \end{cases}

= {\begin{cases} 5 \sqrt{13} + 18 \\ - 5 \sqrt{13} + 18 \end{cases}

\Rightarrow \frac{x^{2} - 1}{x} = \frac{25 \times 13 + 180 \sqrt{13} + 364 - 1}{5 \sqrt{13} + 18} = \frac{180 \sqrt{13} + 648}{5 \sqrt{13} + 18}

= \frac{36 (5 \sqrt{13} + 18)}{5 \sqrt{13} + 18} = 36 .

Commented by Tawa11 last updated on 23/Jul/22

Great sir

Answered by mr W last updated on 22/Jul/22

x - 27 - 9 \sqrt[3]{x} (\sqrt[3]{x} - 3) = x - 36

\sqrt[3]{x} (\sqrt[3]{x} - 3) = 1

\sqrt[3]{x} - \frac{1}{\sqrt[3]{x}} = 3

x - \frac{1}{x} - 3 \times 3 = 27

x - \frac{1}{x} = 36 = \frac{x^{2} - 1}{x}

Commented by Tawa11 last updated on 23/Jul/22

Great sir

Facebook Tweet Pin