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

Let $p(x)=3x^3-5x^2-11x-3$

Then, $p(-1)=3(-1{)}^3-5(-1{)}^2-11(-1)-3$

$=-3-5+11-3$

$=0$

$p(-\frac{1}{3})=3(-\frac{1}{3}{)}^3-5(-\frac{1}{3}{)}^2-11(-\frac{1}{3})-3$

$=(-\frac{1}{9})-(\frac{5}{9})+(\frac{11}{3})-3$

$=(\frac{-1-5+33-27}{9})$

$=0$

$p(3)=3(3{)}^3-5(3{)}^2-11(3)-3$

$=81-45-33-3=0$

Hence, we verified that $3,-1$ and $-\frac{1}{3}$ are the zeroes of the given polynomial.

So, we take $\alpha =3,\beta =-1$ and $\gamma =-\frac{1}{3}$

Verification: 

$\alpha +\beta +\gamma =3+(-1)+(-\frac{1}{3})=(\frac{5}{3})$

$=-\frac{Coefficientof{x}^2}{Coefficientof{x}^3}=-\frac{5}{3}$

$\alpha \beta +\beta \gamma +\gamma \alpha$

$=(3)(-1)+(-1)(-\frac{1}{3})+(-\frac{1}{3})(3)=(-\frac{11}{3})$

$=\frac{Coefficientofx}{Coefficientof{x}^3}=\frac{-11}{3}=(-\frac{11}{3})$

And 

$\alpha \beta \gamma =3\times -1\times (-\frac{1}{3})=1$

$=-\frac{Constantterm}{Coefficientof{x}^3}=-\frac{-3}{3}=1$

Thus, the relationship between the zeroes and the coefficients is verified.