If X = √2 -1 then find the value of x^2 + 1/x^2

1.	If X = √2 -1 then find the value of x^2 + 1/x^2
Answer» Solution:- *Given:- $\rm \: x = \sqrt{2} - 1$ To FIND; $\rm \: {x}^{2} + \frac{1}{ {x}^{2} }$ PUT* *the* *value* of x on *equation* $\rm \: ( \sqrt{2} - 1) {}^{2} + \frac{1}{ ( \sqrt{2} - 1) {}^{2} }$ *Using* *this* *identity* $\rm(a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab$ We *get, $\rm \: ( \sqrt{2} ) {}^{2} + 1 - 2 \sqrt{2} + \frac{1}{ ( \sqrt{2} ) {}^{2} + 1 - 2 \sqrt{2}}$ $\rm \: 3 - 2 \sqrt{2} + \frac{1}{ 3 - 2 \sqrt{2}}$ TAKING* *lcm* $\rm \frac{ (3 - 2 \sqrt{2})( 3 - 2 \sqrt{2} ) + 1}{ 3 - 2 \sqrt{2}}$ $\rm \: \frac{( 3 - 2 \sqrt{2}) {}^{2} + 1 }{ 3 - 2 \sqrt{2}}$ $\rm \: \frac{(3) {}^{2} + (2 \sqrt{2} ) {}^{2} - 2 \times 3 \times 2 \sqrt{2} + 1 }{ 3 - 2 \sqrt{2}}$ $\rm \: \frac{9 + 8 - 12 \sqrt{2} + 1 }{ 3 - 2 \sqrt{2}}$ $\rm \: \frac{18 - 12 \sqrt{2} }{ 3 - 2 \sqrt{2}}$ *Taking* *common* $\rm \: \frac{<klux>6</klux>( 3 - 2 \sqrt{2})}{ 3 - 2 \sqrt{2}}$ *Answer*:- 6

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