If a² - 3a + 1 = 0, find

(i) $a^2 + \dfrac{1}{a^2}$

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

Dividing each term of a² - 3a + 1 = 0 by a we get,

$a + \dfrac{1}{a} = 3$.

(i) We know that,

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

Substituting values we get,

$a^2 + \dfrac{1}{a^2} = 3^2 - 2 = 9 - 2 = 7.$

Hence, $a^2 + \dfrac{1}{a^2}$ = 7.

(ii) We know that,

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

Substituting values we get,

$a^3 + \dfrac{1}{a^3} = 3^3 - 3 \times 3 = 27 - 9 = 18.$

Hence, $a^3 + \dfrac{1}{a^3} = 18.$