TITLE: Find $m,\,n$ so the equation $\left ( * \right )$ has 3 distinct non-zero real roots $a,\,b,\,c$
QUESTION [0 upvotes]: Given the equation:
$$x^{3}+ mx^{2}+ n= 0\left ( * \right )$$
Find $m,\,n$ so the equation $\left ( * \right )$ has 3 distinct non-zero real roots $a,\,b,\,c$ satisfying $$\frac{a^{4}}{a^{3}- 2\,n}+ \frac{b^{4}}{b^{3}- 2\,n}+ \frac{c^{4}}{c^{3}- 2\,n}= 3$$
I try to use the Carnado's method and Wolfram Alpha, I found:
$$a\approx -1,\,4,\,b\approx 2,\,8,\,c\approx 2,\,8,\,m\approx-4,\,2,\,n\approx-10,\,976$$
Thus, we can assume $- 2\,a= b= c$, and replace on the equation. But the expression is too large and I can't continue! Help me! Thanks!

REPLY [1 votes]: You should start writing down Vieta's Formulas, and use these constantly :
$$a+b+c=-m,\quad ab+bc+ca=0,\qquad abc=-n$$
If you now look at a single denominator, you can rewrite it as :
$$ a^3-2n=a^3+2abc = a(a^2+2bc)$$
$$\qquad=a\left(a^2-2(ab+ac)\right)=a^2(3a+2m)$$
So, the complete fraction becomes now :
$$\frac{a^2}{3a+2m}+\frac{b^2}{3b+2m}+\frac{c^2}{3c+2m}=3$$
If you place everything on a single denominator, and you juggle around with Vieta's formulas, you should get something in the following line :
$$-\frac{4 m^4 + 9 m n - 6 m^2 n}{4 m^3 + 27 n}=-\frac{4 m^3 + (9 - 6 m) n}{4 m^3 + 27 n}\cdot m=3$$
Although this has many solutions, a possible integer solution could be obtained by making this division work, i.e: $m=-3$ and any integer $n$ different from zero.
Eg. $x^3-3x^2+2=0$, has zero's $1,1-\sqrt{3},1+\sqrt{3}$ which satisfy this condition.
Furthermore, since you requested only to have real roots, your cubic discriminant should be bigger than zero. In this case, you have $\Delta=(4m^3+27n)n>0$
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