If $x = \dfrac{1}{4 - x}$, find the values of

(i) $x + \dfrac{1}{x}$

(ii) $x^3 + \dfrac{1}{x^3}$

(iii) $x^6 + \dfrac{1}{x^6}$

(i) Given,

$\phantom{\Rightarrow} x = \dfrac{1}{4 - x} \\[1em] \Rightarrow x(4 - x) = 1 \\[1em] \Rightarrow 4x - x^2 = 1 \\[1em]$

On dividing above equation by x,

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

Hence, the value of $x + \dfrac{1}{x}$ = 4.

(ii) We know that,

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

Substituting values we get,

$\Big(x^3 + \dfrac{1}{x^3}\Big) = 4^3 - 3 \times 4 = 64 - 12 = 52.$

Hence, the value of $x^3 + \dfrac{1}{x^3}$ = 52.

(iii) We know,

$\Big(x^6 + \dfrac{1}{x^6}\Big) = \Big(x^3 + \dfrac{1}{x^3}\Big)^2 - 2$.

Substituting values we get,

$\Big(x^6 + \dfrac{1}{x^6}\Big) = 52^2 - 2 = 2704 - 2 = 2702.$

Hence, the value of $x^6 + \dfrac{1}{x^6}$ = 2702.