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.

Alright, I have this problem:

**Problem Statement:**
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 \( x^4 + y^4 + z^4 \).

### Understanding the Problem
First, I need to find the sum of the fourth powers of \( x, y, z \) given the sums of their first, second, and third powers.

### Step 1: Find the Sum of Products \( xy + yz + zx \)
I recall that for any three numbers, the following identity holds:
\[
(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)
\]
I already know \( x + y + z = 6 \) and \( x^2 + y^2 + z^2 = 14 \). Plugging these into the identity:
\[
6^2 = 14 + 2(xy + yz + zx) \\
36 = 14 + 2(xy + yz + zx) \\
36 - 14 = 2(xy + yz + zx) \\
22 = 2(xy + yz + zx) \\
xy + yz + zx = 11
\]

### Step 2: Find the Sum of Products \( xyz \)
I recall another identity related to the sum of cubes:
\[
x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)
\]
I know \( x^3 + y^3 + z^3 = 36 \), \( x + y + z = 6 \), \( x^2 + y^2 + z^2 = 14 \), and \( xy + yz + zx = 11 \). Plugging these in:
\[
36 - 3xyz = 6(14 - 11) \\
36 - 3xyz = 6 \times 3 \\
36 - 3xyz = 18 \\
-3xyz = 18 - 36 \\
-3xyz = -18 \\
xyz = 6
\]

### Step 3: Find the Sum of Fourth Powers \( x^4 + y^4 + z^4 \)
Now, I need to find \( x^4 + y^4 + z^4 \). To do this, I think about using the recurrence relation or Newton's identities.

First, I recall that for any three numbers, the following recurrence relation holds:
\[
x^{n+1} + y^{n+1} + z^{n+1} = (x + y + z)(x^n + y^n + z^n) - (xy + yz + zx)(x^{n-1} + y^{n-1} + z^{n-1}) + xyz(x^{n-2} + y^{n-2} + z^{n-2})
\]

For \( n = 3 \), I can find \( x^4 + y^4 + z^4 \):
\[
x^4 + y^4 + z^4 = (x + y + z)(x^3 + y^3 + z^3) - (xy + yz + zx)(x^2 + y^2 + z^2) + xyz(x + y + z)
\]

I already have all the necessary values:
- \( x + y + z = 6 \)
- \( x^3 + y^3 + z^3 = 36 \)
- \( xy + yz + zx = 11 \)
- \( x^2 + y^2 + z^2 = 14 \)
- \( xyz = 6 \)

Plugging them in:
\[
x^4 + y^4 + z^4 = 6 \times 36 - 11 \times 14 + 6 \times 6 \\
= 216 - 154 + 36 \\
= (216 - 154) + 36 \\
= 62 + 36 \\
= 98
\]

### Verification
Just to make sure, I can think about the roots of a polynomial whose roots are \( x, y, z \). The elementary symmetric polynomials are:
- \( e_1 = x + y + z = 6 \)
- \( e_2 = xy + yz + zx = 11 \)
- \( e_3 = xyz = 6 \)

The monic polynomial is:
\[
P(t) = t^3 - e_1 t^2 + e_2 t - e_3 = t^3 - 6t^2 + 11t - 6
\]

The roots of \( P(t) \) are \( x, y, z \). To find \( x^4 + y^4 + z^4 \), I can use the recurrence relation derived from the polynomial.

For any root \( r \) of \( P(t) \), we have:
\[
r^3 = 6r^2 - 11r + 6
\]

Multiply both sides by \( r \) to get:
\[
r^4 = 6r^3 - 11r^2 + 6r
\]

Now, sum over all roots \( x, y, z \):
\[
x^4 + y^4 + z^4 = 6(x^3 + y^3 + z^3) - 11(x^2 + y^2 + z^2) + 6(x + y + z)
\]

Again, plugging in the known values:
\[
x^4 + y^4 + z^4 = 6 \times 36 - 11 \times 14 + 6 \times 6 = 216 - 154 + 36 = 98
\]

This matches the previous result, so I can be confident in the answer.

### Final Answer
After careful consideration and verification, the value of \( x^4 + y^4 + z^4 \) is:

\[
\boxed{98}
\]