If $x + \dfrac{1}{x} = 2$, prove that $x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}$.

We know that,

$x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2$.

Substituting values we get,

$x^2 + \dfrac{1}{x^2} = 2^2 - 2 = 4 - 2 = 2$ ……..(i)

We know that,

$x^3 + \dfrac{1}{x^3} = \Big(x + \dfrac{1}{x}\Big)^3 - 3\Big(x + \dfrac{1}{x}\Big)$

Substituting values we get,

$x^3 + \dfrac{1}{x^3} = (2)^3 - 3 \times 2 = 8 - 6 = 2$ …….(ii)

We know that,

$x^4 + \dfrac{1}{x^4} = \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - 2$

Substituting values we get,

$x^4 + \dfrac{1}{x^4} = 2^2 - 2 = 4 - 2 = 2$ ………..(iii)

From (i), (ii) and (iii),

Hence proved, $x^2 + \dfrac{1}{x^2} = x^3 + \dfrac{1}{x^3} = x^4 + \dfrac{1}{x^4}.$ when $x + \dfrac{1}{x} = 2$