Question

if x+1/x=2, prove that x^2+1/x^2=x^3+1/x^3=x^4+1/x^4​

Answer 1

Given,

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

$\longrightarrow\sf{\dfrac{x^2+1}{x}=2}$

$\longrightarrow\sf{x^2+1=2x}$

$\longrightarrow\sf{x^2-2x+1=0}$

$\longrightarrow\sf{(x-1)^2=0}$

$\Longrightarrow\sf{x=1}$

Then we must notice that, for some real number n,

$\longrightarrow\sf{x^n+\dfrac{1}{x^n}=1^n+\dfrac{1}{1^n}}$

$\longrightarrow\sf{x^n+\dfrac{1}{x^n}=1+\dfrac{1}{1}}$

$\longrightarrow\sf{x^n+\dfrac{1}{x^n}=1+1}$

$\longrightarrow\sf{x^n+\dfrac{1}{x^n}=2}$

Hence it actually implies that the equation $\sf{x^n+\dfrac{1}{x^n}=2}$ is satisfied for every real number values of n if and only if $\sf{x+\dfrac{1}{x}=2}$ or $\sf{x=1.}$

Thus we can say that, if $\sf{x+\dfrac{1}{x}=2,}$ then,

$\longrightarrow\sf{\underline{\underline{x^2+\dfrac{1}{x^2}=x^3+\dfrac{1}{x^3}=x^4+\dfrac{1}{x^4}=x^5+\dfrac{1}{x^5}=\ldots}}}$

Hence Proved!