<s>[INST] Evaluating $\frac{a^3+b^3+c^3-3abc}{a^4+b^4+c^4-4abc}$, given $a+b+c=a^2+b^2+c^2=1$ How do I solve this problem?
Says so:

If  $$a+b+c=a^2+b^2+c^2=1$$
find the value of
$$M=\frac{a^3+b^3+c^3-3abc}{a^4+b^4+c^4-4abc}$$

edit:
I know these identities
$$1)(a + b + c)^2 = a^2 + + c^2 +  b^2+2 a b + + 2 a c + 2 b c$$
$$2)(a + b + c)^3 =3 a^2 b+3 a^2 c+a^3+3 a b^2+6 a b c+3 a c^2+3 b^2 c+b^3+3 b c^2+c^3$$
$$3)(a + b + c)^4=6 a^2 b^2+12 a^2 b c+4 a^3 b+6 a^2 c^2+4 a^3 c+a^4+12 a b^2 c+4 a b^3+12 a b c^2+4 a c^3+6 b^2 c^2+4 b^3 c+b^4+4 b c^3+c^4$$
 [/INST] Expanding on the comments, given that $\,a+b+c=1\,$ and $\,ab+bc+ca=0\,$:
$$
\begin{align}
p(x) = (x-a)(x-b)(x-c) &= x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc
\\ &= x^3 - x^2 - abc
\end{align}
$$
Since $\,p(a)=p(b)=p(c)=0\,$:
$$
\begin{cases}
a^3 = a^2 + abc
\\ b^3 = b^2 + abc
\\ c^3 = c^2 + abc
\end{cases}
$$
Adding the three equalities:
$$
a^3+b^3+c^3=a^2+b^2+c^2 + 3 abc = 1 + 3 abc \;\;\iff\;\; a^3+b^3+c^3 - 3abc = 1
$$
Multiplying the same equalities by $a,b,c$, respectively, and adding up:
$$
a^4+b^4+c^4 = a^3+b^3+c^3 + abc(a+b+c) = 1 + 4abc \;\;\iff\;\; a^4+b^4+c^4-4abc = 1
$$
Note: first identity is equivalent to Vieta's formulas, the others Newton's identities in disguise.
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