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.
 A: 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.

A: Regarding the requirement that $a,c$ be real.  Set $a = 3\mathrm{i}$, $b = -3$, $c = -\mathrm{i}$.  Then $b < 0$, $ab = -9\mathrm{i} = 9c$, and $f(x) = x^3 + 3 \mathrm{i} x^2 - 3 x - \mathrm{i} = (x+\mathrm{i})^3$, so all three roots are the same and none are real.  Consequently, for the problem to be correct, another contraint is required.  That $a$, $b$, and $c$ must be real is a sufficient such constraint.

If $c = 0$, $ab=9c$ and $b<0$ force $a = 0$, so $f(x) = x^3 + bx = (x^2+b)x$, having the three real roots $\{-\sqrt{-b},0,\sqrt{-b}\}$, which are distinct.
From now on, we assume $c \neq 0$, so $0$ is not a root of $f$.  First we address two cases where the roots are real but have duplicates, showing that if the roots are real, they are distinct.  Then we show that having a conjugate pair of roots is impossible.  
If the roots are real and not distinct, then either they are all the same or exactly two are the same.
Suppose (for purposes of contradiction) the roots are $u$, $u$, and $u$ (and $u \neq 0$).  $f(x) = (x-u)^3= x^3 - 3u x^2 + 3u^2 x - u^3$, so \begin{align*}
a &= -3u  \text{,}  \\
b &= 3u^2  \text{, and}  \\
c &= -u^3  \text{.}
\end{align*}
But this says $b > 0$.  So the three roots cannot all be the same.
Now suppose (for purposes of contradiction) the roots are $u$, $u$, and $v$ (and $u \neq 0$, $v \neq 0$, and $v \neq u$).  $f(x) = (x-u)^2(x-v) = x^3 -(2u+v)x^2 + (u^2 + 2uv)x - u^2 v$, so \begin{align*}
a &= -(2u+v)  \text{,}  \\
b &= u^2 + 2uv  \text{, and}  \\
c &= -u^2 v  \text{.}
\end{align*}
From $ab=9c$, \begin{align*}
-(2u+v)(u^2 + 2uv) &= 9(-u^2 v)  \\
-2u^3 -4 u^2 v - u^2 v - 2 uv^2 &= -9 u^2 v  \\
-2u^3 - 2 uv^2 &= -4 u^2 v  \\
-2u(u^2 - 2uv + v^2) &= 0  \\
-2u(u - v)^2 &= 0  \text{.}
\end{align*}
Since a product is zero only if one (or more) of the multiplicands is zero, either $-2u=0$ or $(u-v)^2 = 0$.  The first says $u = 0$, a contradiction.  The second says $v = u$, a contradiction.  This shows not exactly two roots are the same.
Therefore, the if the roots are real, the three roots of $f$ are distinct.
The coefficients of $f$ are real.  (Assumed additional constraint, see the aside before this solution.)  Consequently, either all three roots are real, or one is real and two are a conjugate pair.  Suppose two are a conjugate pair so the roots are $u$, $v + \mathrm{i}w$, $v - \mathrm{i}w$.  By the above, we may take $u \neq 0$ and $w > 0$.  Then 
$f(x) = (x-u)(x-v - \mathrm{i}w)(x - v + \mathrm{i}w) = x^3 - (u+2v)x^2 + (2uv+v^2+w^2)x - (uv^2+uw^2)$, so \begin{align*}
a &= - (u+2v)  \text{,}  \\
b &= (2uv+v^2+w^2)  \text{, and}  \\
c &= - (uv^2+uw^2)  \text{.}
\end{align*}
If $v = 0$, from $ab=9c$,$-(u)(w^2) = -9(uw^2)$, so $8 uw^2 = 0$ and either $u=0$ (contradiction) or $w = 0$ (contradiction).  Therefore, $v \neq 0$.
From $ab=9c$, \begin{align*}
-(u+2v)(2uv+v^2+w^2) &= -9(uv^2+uw^2)  \\
-2u^2v -u v^2 -u w^2 - 4uv^2-2v^3-2vw^2 &= -9uv^2 -9uw^2  \\
-2u^2v -2v^3-2vw^2 &= -4uv^2 -8uw^2  \\
(-2v)u^2 +(4v^2 +8w^2)u -2v(v^2+w^2) &= 0  \text{.}
\end{align*}
By the quadratic formula, $u = \frac{1}{v}\left( v^2 + 2w^2 \pm w\sqrt{3v^2+4w^2} \right)$.  The only constraint we haven't used is $b < 0$, so we substitute these into the expression for $b$ and simplify to $b = 3v^2 + 5w^2 \pm 2w\sqrt{3v^2 + 4w^2}$.  If we pick the positive root, each term is positive, so we must take the negative root and have $2w\sqrt{3v^2 + 4w^2} > 3v^2 + 5w^2$.  Both sides are positive, so squaring only risks introducing negative solutions (of no interest to us), giving \begin{align*}
12v^2w^2 + 16 w^4 &> 9v^4 + 30 v^2w^2 + 25w^4  \\
0 &> 9v^4 + 18 v^2w^2 + 9w^4  \text{,}
\end{align*}
an impossibility.  Therefore, it is not the case that $f$ has a conjugate pair of roots.
Therefore, there are three distinct roots, all of which are real.
