Question

Chapter - Laws of Indices

If $x=(\sqrt{2}-1)^{\frac{1}{3}}$ , then prove that

$(x-\frac{1}{x})^{3}+3(x-\frac{1}{x})+2=0$​

Answer 1

$x=(\sqrt{2}-1)^{\frac{1}{3}}$
$(x-\frac{1}{x})^{3}+3(x-\frac{1}{x})+2\\ {x}^{3} - \frac{1}{ {x}^{3} } + 2 \\ {(\sqrt{2}-1)^{\frac{1}{3} \times 3}}^{} - \frac{1}{ (\sqrt{2}-1)^{\frac{1}{3} \times 3}} + 2 \\ \sqrt{2} - 1 - \frac{1}{ \sqrt{2} - 1} + 2 \\ \frac{ {( \sqrt{2} - 1)}^{2} - 1 }{ \sqrt{2} - 1} + 2 \\ \frac{2 - 2 \sqrt{2} + 1 - 1 + 2 \sqrt{2} - 2}{ \sqrt{2} - 1} = \frac{0}{ \sqrt{2} - 1 } = 0$

Answer 2

$\LARGE{\red{\boxed{\purple{\underline{\orange{\mathtt{Question:↓}}}}}}}$

If $x=(\sqrt{2}-1)^{\frac{1}{3}}$ , then prove that

$(x-\frac{1}{x})^{3}+3(x-\frac{1}{x})+2=0$

$\LARGE{\red{\boxed{\purple{\underline{\orange{\mathtt{Solution:↓}}}}}}}$

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

$\begin{gathered}(x-\frac{1}{x})^{3}+3(x-\frac{1}{x})+2\\ {x}^{3} - \frac{1}{ {x}^{3} } + 2 \\ {(\sqrt{2}-1)^{\frac{1}{3} \times 3}}^{} - \frac{1}{ (\sqrt{2}-1)^{\frac{1}{3} \times 3}} + 2 \\ \sqrt{2} - 1 - \frac{1}{ \sqrt{2} - 1} + 2 \\ \frac{ {( \sqrt{2} - 1)}^{2} - 1 }{ \sqrt{2} - 1} + 2 \\ \frac{2 - 2 \sqrt{2} + 1 - 1 + 2 \sqrt{2} - 2}{ \sqrt{2} - 1} = \frac{0}{ \sqrt{2} - 1 } = 0\end{gathered}$

Hope it helps you.