If x ≠ 0 and $x + \dfrac{1}{x} = 2$; then show that :

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

By formula,

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

Substituting values we get :

$\Rightarrow x^2 + \dfrac{1}{x^2} = 2^2 - 2 \\[1em] = 4 - 2 \\[1em] = 2.$

By formula,

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

Substituting values we get :

$\Rightarrow x^3 + \dfrac{1}{x^3} = 2^3 - 3 \times 2 \\[1em] = 8 - 6 \\[1em] = 2.$

By formula,

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

Substituting values we get :

$\Rightarrow x^4 + \dfrac{1}{x^4} = 2^2 - 2 \\[1em] = 4 - 2 \\[1em] = 2.$

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