Question

x= 1+ root 2. find (x-1/x)³
I m in grade 9​

Answer 1

Answer:

2√2 / (5√2 + 7)

Step-by-step explanation:

Given :

• x = 1 + √2

Expanding

• (x - 1 / x)³
• (1 + √2 - 1)³ / (1 + √2)³
• (√2)³ / (1 + √2)³
• 2√2 / 1 + 2√2 + 3(1)(√2)(1 + √2)
• 2√2 / 1 + 2√2 + 3√2 + 6
• 2√2 / (5√2 + 7)

Answer 2

Answer:

$-14\sqrt{2}+20$

Step-by-step explanation:

$\textsf{Substituting}\quad x=1+\sqrt2\quad\textsf{into}\quad\left(\dfrac{x-1}{x}\right)^3$

$\implies \left(\dfrac{(1+\sqrt2)-1}{(1+\sqrt2)}\right)^3$

$\implies \left(\dfrac{\sqrt2}{1+\sqrt2}\right)^3$

$\textsf{Apply exponent rule}:\left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}$

$\implies \dfrac{(\sqrt2)^3}{(1+\sqrt2)^3}$

Expand numerator:

$(\sqrt{2} )^3=\sqrt{2}\sqrt{2}\sqrt{2}=2\sqrt{2}$

Expand denominator:

$\begin{aligned}(1+\sqrt2)^3 & = (1+\sqrt2)(1+\sqrt2)(1+\sqrt2)\\ & =(1+\sqrt2)(1+2\sqrt{2}+2)\\ & =1+2\sqrt{2}+2+\sqrt{2}+2\sqrt{2}\sqrt{2}+2\sqrt{2}\\ & = 7+5\sqrt{2}\end{aligned}$

$\implies \dfrac{(\sqrt2)^3}{7+5\sqrt2}$

Substituting expanded numerator and denominator:

$\implies \dfrac{2\sqrt{2}}{7+5\sqrt2}$

$\textsf{Mulitply by conjugate :}\quad\dfrac{7-5\sqrt2}{7-5\sqrt2}$

$\implies \dfrac{2\sqrt{2}(7-5\sqrt2)}{(7+5\sqrt2)(7-5\sqrt2)}$

$\implies \dfrac{14\sqrt{2}-20}{49-50}$

$\implies \dfrac{14\sqrt{2}-20}{-1}$

$\implies -14\sqrt{2}+20$