If the roots of the cubic equation $ax^3+bx^2+cx+d=0$ are equal, can one then establish a relationship between $a, b, c, d$?

If the root is triple, then the polynomial is $a(x-r)^3$. This means that $r=-b/3a$ and you can conclude $$ c=\frac{b^2}{3a}\quad d=\frac{b^3}{27a^2}.$$

Using Vieta's formulas we get relations as $$b^3=27a^2d$$ $$b^2=3ac$$ And $$bc=9ad$$

Then note that if the root of the cubic is $\alpha$ with multiplicity $3$, we have: $$(x-\alpha) ^3=0$$ $$\implies \color{red}{1}x^3\color{green}{-3\alpha}x^2\color{blue}{+3\alpha^2}x\color{orange}{-\alpha^3} =0$$ Compare this with the polynomial form you have in mind.