math-algebra-01

# Power Sums of a Symmetric System

Real numbers $x$, $y$, $z$ satisfy the system

$$
\begin{aligned}
x + y + z &= 6,\\
x^2 + y^2 + z^2 &= 14,\\
x^3 + y^3 + z^3 &= 36.
\end{aligned}
$$

Find the value of $x^4 + y^4 + z^4$.

Give your final answer as a single integer.

To find \( x^4 + y^4 + z^4 \), we start by determining the elementary symmetric sums \( s_1 \), \( s_2 \), and \( s_3 \) using the given equations. 

1. From \( x + y + z = 6 \), we have \( s_1 = 6 \).
2. Using \( x^2 + y^2 + z^2 = 14 \) and the identity \( x^2 + y^2 + z^2 = s_1^2 - 2s_2 \), we find \( s_2 = 11 \).
3. Using \( x^3 + y^3 + z^3 = 36 \) and the identity \( x^3 + y^3 + z^3 = s_1^3 - 3s_1s_2 + 3s_3 \), we find \( s_3 = 6 \).

Next, we use the identity \( x^4 + y^4 + z^4 = (x^2 + y^2 + z^2)^2 - 2(x^2y^2 + y^2z^2 + z^2x^2) \). 

4. We calculate \( x^2y^2 + y^2z^2 + z^2x^2 \) using \( (xy + yz + zx)^2 = x^2y^2 + y^2z^2 + z^2x^2 + 2xyz(x + y + z) \), which gives \( x^2y^2 + y^2z^2 + z^2x^2 = 49 \).

Finally, substituting into the identity:

\[
x^4 + y^4 + z^4 = 14^2 - 2 \times 49 = 196 - 98 = 98
\]

The value of \( x^4 + y^4 + z^4 \) is \(\boxed{98}\).