Prove that polynomial $x^3+ax^2+17x+3b$ doesn't have same roots. All roots of polynomial $x^3+ax^2+17x+3b$, $a,b\in \Bbb Z$ are integers. Prove that this polynomial doesn't have same roots.


My plan was to find all three solutions and then compare them.

I don't know how to find the first root - tried to use cubic formula to find it but I got a large expression, just a few numbers canceled.  Any advices? Thanks.
Say it has, so $x_1=x_2 = m$ and $x_3=k$, then we have by (second) Vieta formula $$m^2+2mk =17$$ and (third) $$m^2k =-3b$$

So $m$ or $k$ is divisible by $3$. Clearly $m$ is not (since else $3\mid 17$), so $3\mid k$. But then we have $m^2\equiv 2\pmod 3$, a contradiction.
Hint $ $ the roots are distinct $\bmod 3\!:\ f \equiv x(x^2\!+\!ax\!-1)\not \equiv x(x\!-\!r)^2\,$ by $\,r^2\not\equiv 1$