Show that this polynomial has three different real roots There is polynomial $f(x)=x^3 +ax^2+bx+c$ and $ab=9c$ and $b<0$. Like in the title. I've only come up so far with $9x^3 + 9ax^2 + 9bx+ab=0$ and factoring here does nothing.

I have expanded Eric Towers' comments into an answer.  Hope this can help OP.  The second method is more efficient and error-prone doesn't work well, unlike the first one.
Method 1: Discriminant
Given that $ab=9c$ and $b<0$, the discriminant is
\begin{align}
\Delta &= 18(1)abc-4a^3c+a^2b^2-4(1)b^3-27(1)^2c^2 \\
&= 162c^2-4a^3c+81c^2-4b^3-27c^2 \\
&= 216c^2-4a^3c-4b^3 \\
&= 4\left(54c^2-a^3c-b^3\right) \\
&= 4\left[54c^2-\left(\frac{9c}{b}\right)^3c-b^3\right] \\
&= 4\left[54c^2+\frac{3^6c^4}{(-b)^3}+(-b)^3\right] \\
&= \frac{4}{(-b)^3} \left[\left(3^3c^2\right)^2+2\left(3^3c^2\right)(-b)^3+(-b)^6\right] \\
&= \frac{4}{\underbrace{(-b)^3}_{b<0}} (\underbrace{27c^2}_{\ge0}+\underbrace{(-b)^3}_{b<0})^2 > 0
\end{align}
Hence $f(x)=x^3 +ax^2+bx+c$ has three distinct real roots.
Method 2: Descartes rule of signs
Edit: As pointed out in the comments, this argument is incomplete.  I tried proof by contradiction, and dividing by $x-r$ ($r$ is the positive root) in the case $a>0$, but I can't figure out how this works.
I'm leaving this, hoping some motivated user complete this
Refer to Wiki's example to know how to apply this result.
Rewrite $f(x)=x^3 +ax^2+bx+c$ into an array of $\pm$: $(+,?,-,?)$.  Since $ab=9c$ with $b<0$, the first and the third $?$'s are opposite to each other.
\begin{array}{c|c|c}
\text{Case} & \text{+ve roots} & \text{-ve roots} \\ \hline
a>0 & +,+,-,- & -,+,+,- \\
 & \text{1 sign change} & \text{2 sign changes} \\ \hline
a<0 & +,-,-,+ & -,-,+,+ \\
 & \text{2 sign changes} & \text{1 sign change} \\ \hline
a=0 & +,\phantom{+},-,\phantom{+} & -,\phantom{+},+,\phantom{+} \\
 & \text{1 sign change} & \text{1 sign change}
\end{array}


*

*Divide the situation into three cases according to the sign of $a$.

*Extract the signs from $f(x)$ and write them down in the first column.

*Count the number of sign changes ($+,-$ or $-,+$) in each cell.  This represents the parity of the number of positive root(s) in each case.

*Write down the signs for $f(-x)$ in the second column.


*

*Copy the signs representing even powers of $x$.

*Invert the signs representing odd powers of $x$.


*Repeat (3) in the second column.  This represents the parity of the number of negative root(s) in each case.





*

*If $a\ne0$, we can't conclude anything from it.

*If $a=0$, the condition $ab=9c$ gives $c=0$, so $x=0$ is a root not counted by this method.  We have three distinct real roots.

