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.

I need to find the value of $x^4 + y^4 + z^4$ given the system of equations involving power sums.

Let me denote:
- $s_1 = x + y + z = 6$
- $s_2 = x^2 + y^2 + z^2 = 14$
- $s_3 = x^3 + y^3 + z^3 = 36$
- $s_4 = x^4 + y^4 + z^4$ (what I want to find)

I'll use Newton's identities, which relate power sums to elementary symmetric polynomials.

First, let me find the elementary symmetric polynomials:
- $e_1 = x + y + z = 6$
- $e_2 = xy + yz + zx$
- $e_3 = xyz$

From the identity $(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$:
$$6^2 = 14 + 2(xy + yz + zx)$$
$$36 = 14 + 2(xy + yz + zx)$$
$$xy + yz + zx = 11$$

So $e_2 = 11$.

Now I'll use the identity for power sums in terms of elementary symmetric polynomials:
$$s_3 = e_1 s_2 - e_2 s_1 + 3e_3$$

Substituting known values:
$$36 = 6 \cdot 14 - 11 \cdot 6 + 3e_3$$
$$36 = 84 - 66 + 3e_3$$
$$36 = 18 + 3e_3$$
$$3e_3 = 18$$
$$e_3 = 6$$

Now I'll find $s_4$ using the recurrence relation:
$$s_4 = e_1 s_3 - e_2 s_2 + e_3 s_1$$

Substituting all known values:
$$s_4 = 6 \cdot 36 - 11 \cdot 14 + 6 \cdot 6$$
$$s_4 = 216 - 154 + 36$$
$$s_4 = 98$$

Let me verify this is correct by checking if there's a consistent system.

I can also verify using another approach. The polynomial whose roots are $x$, $y$, $z$ is:
$$t^3 - e_1 t^2 + e_2 t - e_3 = 0$$
$$t^3 - 6t^2 + 11t - 6 = 0$$

Let me check if this polynomial has real roots that satisfy our conditions.

Factoring: $t^3 - 6t^2 + 11t - 6 = (t-1)(t-2)(t-3)$

So the roots are $x = 1$, $y = 2$, $z = 3$.

Checking:
- $x + y + z = 1 + 2 + 3 = 6$ ✓
- $x^2 + y^2 + z^2 = 1 + 4 + 9 = 14$ ✓
- $x^3 + y^3 + z^3 = 1 + 8 + 27 = 36$ ✓

Therefore:
$$x^4 + y^4 + z^4 = 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98$$

$\boxed{98}$