Question

a=1-root2, find the value of (a-1/a)^3

Answer 1

refer the image for solution.

Answer 2

The value of ($\frac{a-1}{a}$)³ is 20 + 14√2

Given,

a = 1 - √2

To Find,

($\frac{a-1}{a}$)³

Solution,

($\frac{a-1}{a}$)³ can be calculated as follows -

a - 1 = 1 - √2 - 1

a - 1 = - √2

($\frac{a-1}{a}$) = $\frac{-\sqrt{2} }{1-\sqrt{2} }$

($\frac{a-1}{a}$)³ = ($\frac{a-1}{a}$) * ($\frac{a-1}{a}$) * ($\frac{a-1}{a}$)

($\frac{a-1}{a}$)³ = $\frac{-\sqrt{2} }{1-\sqrt{2} }$  *  ($\frac{-\sqrt{2} }{1-\sqrt{2} }$) * ($\frac{-\sqrt{2} }{1-\sqrt{2} }$)

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

($\frac{a-1}{a}$)³ = $\frac{-2\sqrt{2} }{1-\sqrt{2} (1+2-2\sqrt{2}) }$  

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

($\frac{a-1}{a}$)³ = $\frac{-2\sqrt{2} }{3-3\sqrt{2}-2\sqrt{2} +4 }$

($\frac{a-1}{a}$)³ = $\frac{-2\sqrt{2} }{7-5\sqrt{2} }$

Rationalizing -

($\frac{a-1}{a}$)³ = $\frac{-2\sqrt{2} }{7-5\sqrt{2} }$ * $\frac{7+5\sqrt{2} }{7+5\sqrt{2} }$

($\frac{a-1}{a}$)³ = $\frac{(-2\sqrt{2})(7+5\sqrt{2} ) }{7^2-(5\sqrt{2})^2 }$

($\frac{a-1}{a}$)³ = $\frac{-20-14\sqrt{2} }{49 - 50 }$

($\frac{a-1}{a}$)³ = $\frac{-20-14\sqrt{2} }{-1 }$

($\frac{a-1}{a}$)³ = 20 + 14√2

Thus, the answer is 20 + 14√2.

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