Verify that $3, -1, -\frac{1}{3}$ are the zeroes of the cubic polynomial $p(x) = 3x^3 - 5x^2 - 11x - 3$,and then verify the relationship between the zeroes and the coefficients.

(A) Comparing the given polynomial $p(x) = 3x^3 - 5x^2 - 11x - 3$ with the standard form $ax^3 + bx^2 + cx + d$,we get $a = 3, b = -5, c = -11, d = -3$.
First,we verify the zeroes:
$p(3) = 3(3)^3 - 5(3)^2 - 11(3) - 3 = 81 - 45 - 33 - 3 = 0$
$p(-1) = 3(-1)^3 - 5(-1)^2 - 11(-1) - 3 = -3 - 5 + 11 - 3 = 0$
$p(-\frac{1}{3}) = 3(-\frac{1}{3})^3 - 5(-\frac{1}{3})^2 - 11(-\frac{1}{3}) - 3 = -\frac{1}{9} - \frac{5}{9} + \frac{11}{3} - 3 = -\frac{6}{9} + \frac{33}{9} - \frac{27}{9} = 0$
Thus,$3, -1, -\frac{1}{3}$ are the zeroes.
Let $\alpha = 3, \beta = -1, \gamma = -\frac{1}{3}$.
$1$. Sum of zeroes: $\alpha + \beta + \gamma = 3 - 1 - \frac{1}{3} = 2 - \frac{1}{3} = \frac{5}{3} = -\frac{b}{a}$.
$2$. Sum of product of zeroes taken two at a time: $\alpha\beta + \beta\gamma + \gamma\alpha = (3)(-1) + (-1)(-\frac{1}{3}) + (-\frac{1}{3})(3) = -3 + \frac{1}{3} - 1 = -4 + \frac{1}{3} = -\frac{11}{3} = \frac{c}{a}$.
$3$. Product of zeroes: $\alpha\beta\gamma = (3)(-1)(-\frac{1}{3}) = 1 = -\frac{d}{a} = -(\frac{-3}{3}) = 1$.
All relationships are verified.