Question

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

Answer 1

2√2 / (5√2 + 7)

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

`-14√(2)+20`

Explanation:

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

`\implies \left(((1+\sqrt2)-1)/((1+\sqrt2))\right)^3`

`\implies \left((\sqrt2)/(1+\sqrt2)\right)^3`

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

`\implies ((\sqrt2)^3)/((1+\sqrt2)^3)`

Expand numerator:

`(√(2) )^3=√(2)√(2)√(2)=2√(2)`

Expand denominator:

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

`\implies ((\sqrt2)^3)/(7+5\sqrt2)`

Substituting expanded numerator and denominator:

`\implies (2√(2))/(7+5\sqrt2)`

`\textsf{Mulitply by conjugate :}\quad(7-5\sqrt2)/(7-5\sqrt2)`

`\implies (2√(2)(7-5\sqrt2))/((7+5\sqrt2)(7-5\sqrt2))`

`\implies (14√(2)-20)/(49-50)`

`\implies (14√(2)-20)/(-1)`

`\implies -14√(2)+20`