Given that $3$, $-1$, and $-\dfrac{1}{3}$ are zeroes of the cubic polynomial $p(x)=3x^3-5x^2-11x-3$, which relation is verified by their product?

Here $\alpha=3$, $\beta=-1$, and $\gamma=-\dfrac{1}{3}$. Their product is $3\times(-1)\times\left(-\dfrac{1}{3}\right)=1$. For a cubic polynomial $ax^3+bx^2+cx+d$, the product of zeroes is $-\dfrac{d}{a}$. Since $a=3$ and $d=-3$, we get $-\dfrac{-3}{3}=1$, so the relation is verified. The other options use the wrong coefficient relation: $-\dfrac{b}{a}$ gives the sum of zeroes, and $\dfrac{c}{a}$ gives the sum of pairwise products. Also, $\dfrac{d}{a}$ has the wrong sign for the product of zeroes of a cubic.