Cyclic Algebra. Suppose that $a,b,c$ are real numbers satisfying $a^2 + b^2 + c^2 = 1$ and $a^3 + b^3 + c^3 = 1$.
Find all possible value(s) of $a+b+c$.
My solution :
$(a+b+c)(a^2+b^2+c^2)=a^3+b^3+c^3+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow   a+b+c=1+a^2b+a^2c+b^2a+c^2a+b^2c+c^2b$
$\Rightarrow a+b+c=1+c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)$
$\Rightarrow   a+b+c=1+c(1-c^2)+a(1-a^2)+b(1-b^2)$
$\Rightarrow   a+b+c=1+c-c^3+a-a^3+b-b^3$
$\Rightarrow   a+b+c=1+c+a+b-1$
$\Rightarrow   a+b+c=c+a+b$
This led me to the same expression $a+b+c$.
Is there a smarter way to solve this problem?

Hint: consider $$(a+b+c)^2$$ and $$(a+b+c)^3$$
we have
$$\frac{a+b+c}{3}\le \sqrt{\frac{a^2+b^2+c^2}{3}}$$
