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Given the cubic equation:
$x^3 - px^2 + qx - r = 0$

Since $a, b, c$ are the roots of the equation, by Vieta's formulas:

1. $a + b + c = p$
2. $ab + bc + ca = q$
3. $abc = r$

(i) Value of $a^2 + b^2 + c^2$:
Using the algebraic identity:
$a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)$
Substituting the values:
$a^2 + b^2 + c^2 = p^2 - 2q$

(ii) Value of $a^3 + b^3 + c^3$:
Using the identity:
$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$a^3 + b^3 + c^3 - 3abc = (a + b + c)[(a + b + c)^2 - 3(ab + bc + ca)]$

Substituting the values:
$a^3 + b^3 + c^3 - 3r = p(p^2 - 3q)$
$a^3 + b^3 + c^3 = p(p^2 - 3q) + 3r$
$a^3 + b^3 + c^3 = p^3 - 3pq + 3r$

Answers:

• $a^2 + b^2 + c^2 = p^2 - 2q$
• $a^3 + b^3 + c^3 = p^3 - 3pq + 3r$