$ f(x)=x^3+ax^2+bx+\frac{ab}{9}$ Let $ b=-d$ with $ d>0$ $ f(x)=x^3+ax^2-dx-\frac{ad}{9}$ If $ a=0$ it is very easy the proof. Case 1: If $ a>0$ $ f(-a)=8ad$ $ f(0)=-ad$ $ f(\sqrt{d})=8ad$ Then $ f(x)$ has at least two different real roots, then it has three real roots. Case2: If $ a>0$ $ f(-a)=8ad$ $ f(0)=-ad$ $ f(-\sqrt{d})=8ad$ Then $ f(x)$ has at least two different real roots, then it has three real roots. We only have two prove that if it has two different real roots, then it has three different real roots. By contradiction, suppose that it has two equal roots. Then: $ f(x)=(x-r)^2(x-t)$ By Viette`s formulas: $ a=-(2r+t)$ $ b=r^2+2tr$ $ ab=-9r^2c$ $ \implies -(2r+t)r(r+2t)=-9r^2t$ $ \implies (r-t)^2=0$ $ \implies r=t$ Which is an contradiction because $ f$ has at least two different roots.