<s>[INST] Solving Yoshida equations I want to solve $a$, $b$ and $c$ out of the following set of equations
\begin{cases}
a + b + c = 1 \\
a^{p+1} + b^{p+1} + c^{p+1} = 0 \\
a = c \\
\end{cases}
where $p$ is even. But I absolutely have no idea how to even start solving it...
Edit: based on the hint given below, the solution goes as follows
\begin{cases}
 2a + b = 1 \\
 2a^{p+1} + b^{p+1} = 0 \\
\end{cases}
because $a=c$. By writing $b = 1-2a$ and substituting this into the second equation, we get $2a^{p+1} + (1-2a)^{p+1} = 0$. Because $p$ is even, this becomes $2a^{p+1} = (2a-1)^{p+1}$. Now taking the $(p+1)th$ root of both sides we obtain
$$ 
a = \frac{1}{2-   \sqrt[\leftroot{-2}\uproot{2}p+1]{2}   }
$$
 [/INST] Hint
So, you have (since $a=c$)
$$\begin{cases}
2a + b= 1, \\
2a^{p+1} + b^{p+1} = 0 ,
\end{cases}$$
from where $b=1-2a$ and then $2a^{p+1}=-(1-2a)^{p+1}.$ Since $p$ is even, we have $\sqrt[p+1]{2}a=2a-1.$
Can you continue from here?
</s>