Question

what is the value of (x + 1/x)^2

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

Explanation:

Concept

two same value or two value having same power or two like terms gets cancelled and the resultant would be 1

$\bold{\red{( { \frac{x + 1}{x}})^{2} }}$

Here,a identity is used :-

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

$= > ( { \frac{x + 1}{x} })^{2} = \frac{ {x}^{2} + {(1)}^{2} + 2x}{{x}^{2} }$

$= > \frac{ {x}^{2} }{ {x}^{2} } + \frac{1}{ {x}^{2} } + \frac{2x}{ {x}^{2} }$

$\bold{ = > 1 + \frac{1}{ {x}^{2} } + \frac{2}{x} = \frac{1}{ {x}^{2} } + \frac{2}{x } + 1}$

Other related identies:

$\bold{\boxed{ {(x -y)}^{2} = {x}^{2} + {y}^{2} - 2xy}}$

$\bold{\boxed{ {x}^{3} + {y}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}}$

$\bold{\boxed{(x + y)(x + z) = {x}^{2} + (y+ z)x + yz}}$

$\bold{\boxed{ {(x +y)}^{3} = {x}^{3} + {y}^{3} +3xy[x+y]}}$