Question

x+1/x=2 then show that!
X²+1x^2 = X^3+1/X3 = x^4+1x^4​

Answer 1

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

to show :

${x}^{2} + \dfrac{1}{ {x}^{2} } = {x}^{3} + \dfrac{1}{ {x}^{3} } = {x}^{4} + \dfrac{1}{ {x}^{4} }$

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Identity to use :

(a+b)² = a² + 2ab + b²

and (a+b)³ = a³ + b³ + 3ab(a+b)

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

x and 1/x gets cancelled

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

$4 = {x}^{2} + 2 + \frac{1}{ {x}^{2} }$

by transposing 2 to the LHS(Left Hand Side)

$4 - 2 = {x}^{2} + \frac{1}{ {x}^{2} }$

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

again using the identity

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

${2}^{2} = {x}^{4} + 2 + \frac{1}{ {x}^{4} }$

$4 = {x}^{4} + 2 + \frac{1}{ {x}^{4} }$

$4 - 2 = {x}^{4} + \frac{1}{ {x}^{4} }$

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Using (a+b)³ = a³ + 3ab(a+b) + b³

$(x + \frac{1}{x})^{3} = {x}^{3} + 3(x)( \frac{1}{x})(x + \frac{1}{x}) + \frac{1}{ {x}^{3} }$

${2}^{3} = {x}^{3} + 3(2) + \frac{1}{ {x}^{3} }$

$8 = {x}^{3} + 6 + \frac{1}{ {x}^{3} }$

$8 - 6 = {x}^{3} + \frac{1}{ {x}^{3} }$

$2 = {x}^{3} + \frac{1}{ {x}^{3} }$

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Therefore Verified.

Hope it helps!

have a great day!