Solving equations $a+b+c=5,a^2+b^2+c^2=11,a^3+b^3+c^3=27$ 
Let $a, b, c $ be real numbers such that $a<b<c$ and satisfying
  $$a+b+c=5;$$
  $$a^2+b^2+c^2=11;$$
  $$a^3+b^3+c^3=27.$$
  Prove that $0<a<1<b<2<c<3$.

My understanding :
$(a+b+c)^2 - (a^2+b^2+c^2) = 14$
so, $ab+bc+ca = 7$
$a^2+b^2+c^2-ab-bc-ca = 4$
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=20$
so, $abc = -\frac{7}{3}$
Thus $a, b, c$ are roots of the equation $P(x), x^3-5x^2+7x-\frac{7}{3}=0$.
We have,
$P(0) = -\frac{7}{3} <0$
$P(1) = 1-5+7-\frac{7}{3} >0$
$P(2) = 8-20+14-\frac{7}{3} <0$
$P(3) = 27-45+21-\frac{7}{3} >0$.
by Intermediate value theorem, $P(x)$ has a root in each of the interval $(0,1), (1, 2), (2,3)$.
Since $a<b<c$, we get $0<a<1<b<2<c<3$.
 A: $$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$
$$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$$
Get from here $ab+ac+bc$ and $abc$.
I got $ab+ac+bc=7$ and $abc=\frac{7}{3}$, which gives that $a$, $b$ and $c$ are roots of the equation
$$x^3-5x^2+7x-\frac{7}{3}=0$$ or
$$\frac{1}{(x-1)^3}-\frac{3}{x-1}=-\frac{3}{2}$$
Now, let $\frac{1}{x-1}=2\cos\phi$.
Hence, $\cos3\phi=-\frac{3}{4}$ and the rest is smooth for  finding of values $a$, $b$ and $c$, but it's unnecessary.
Let $f(x)=x^3-5x^2+7x-\frac{7}{3}$ and check $f(0)$, $f(1)$, $f(2)$ and $f(3)$.
A: This is a classical exercise about symmetric functions. By Newton's identities
$$ e_1 = p_1 = 5, $$
$$ 2e_2 = p_1^2 - p_2 = 14, $$
$$ 3e_3 = e_2 p_1 - e_1 p_2 + p_3 = 7 $$
hence $a,b,c$ are roots of the polynomial
$$ p(z)=(z-a)(z-b)(z-c) = z^3-e_1 z^2+e_2 z-e_3 = z^3-5z^2+7z-\frac{7}{3} $$
and the claim follows from the fact that $p(0)<0, p(1)>0, p(2)<0, p(3)>0$.
Numerically the roots lie at
$ \{0.485131\ldots,1.72235\ldots,2.79252\ldots\} $.

To perform the above manipulations in a single step, one may exploit the $\text{EXP}/\text{LOG}$ map explained here and compute the Taylor series at the origin, up to the cubic term, of
$$ \exp\left(-5z+\frac{11}{2}z^2-9z^3\right)=1+5z+7z^2+\frac{7}{3}z^3+\ldots$$
to recover the coefficients of $p(z)$.
