Question

If x = 1 - V2.find the value of (x-1/x)^3​

Answer 1

Correct Question :

If x = 1 - √2,Find the value of (x-1/x)³

Solution :

$\sf \: x = 1 - \sqrt{2}$

$\longrightarrow \: \sf \dfrac{1}{x} \: = \dfrac{1}{1 - \sqrt{2} }$

Rationalize the denominator :

$\longrightarrow \: \sf \dfrac{1}{x} \: = \dfrac{1 \times 1 + \sqrt{2} }{(1 - \sqrt{2})(1 + \sqrt{2} ) }$

$\longrightarrow \: \sf \dfrac{1}{x} \: = \dfrac{1 + \sqrt{2} }{( {1)}^{2} - (\sqrt{ {2})^{2} } }$

$\longrightarrow \sf \dfrac{1}{x} \: = \dfrac{1 + \sqrt{2} }{1 - 2 }$

$\longrightarrow \sf \dfrac{1}{x} \: = \dfrac{1 + \sqrt{2} }{ -1 }$

$\longrightarrow \: \sf \dfrac{1}{x} \: = - 1 - \sqrt{2}$

Put the value of x and 1/x :

$\bigstar \boxed{\sf \: (x - \frac{1}{x} ) ^{3} }$

$\longrightarrow \: \sf [(1 - \sqrt{2}) - ( - 1 - \sqrt{2} )] ^{3}$

$\longrightarrow \: \sf (1 - \sqrt{2} + 1 + \sqrt{2} ) ^{3}$

$\longrightarrow \: \sf (2) ^{3}$

$\longrightarrow \red{\sf \: 8}$

Answer 2

__________________________

$\huge\tt{GIVEN:}$

• x = 1 - √2

__________________________

$\huge\tt{TO~FIND:}$

• value of (x-1/x)³

__________________________

$\huge\tt{SOLUTION:}$

↪x = 1-√2

↪1/x = 1/(1-√2)

↪1/x = {1×1+√2} / {(1-√2)(1+√2)}

↪1/x = {1+√2} / {1²-√2²}

↪1/x = {1+√2} / {1-2}

↪1/x = {1+√2} / {-1}

↪1/x = -1 - √2

Putting values,

↪[{1-√2} - {-1 -√2} ]³

↪{1-√2+1+√2}³

↪2³

↪8

__________________________