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:

**Given:**
1. \( x + y + z = 6 \)
2. \( x^2 + y^2 + z^2 = 14 \)
3. \( x^3 + y^3 + z^3 = 36 \)

**Find:** \( x^4 + y^4 + z^4 \)

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

### Step 1: Use Symmetric Sums and Newton's Identities
I recall that for any three numbers \( x, y, z \), the elementary symmetric sums are:
- \( S_1 = x + y + z \)
- \( S_2 = xy + yz + zx \)
- \( S_3 = xyz \)

But I have power sums \( P_k = x^k + y^k + z^k \) for \( k = 1, 2, 3 \). I need to find \( P_4 \).

### Step 2: Find \( S_2 \) Using \( P_1 \) and \( P_2 \)
I know the identity:
\[ P_2 = (x + y + z)^2 - 2(xy + yz + zx) \]

Given \( P_1 = 6 \) and \( P_2 = 14 \), I can plug them in:
\[ 14 = 6^2 - 2S_2 \]
\[ 14 = 36 - 2S_2 \]
\[ 2S_2 = 36 - 14 = 22 \]
\[ S_2 = 11 \]

### Step 3: Find \( S_3 \) Using \( P_3 \) and Newton's Identities
I recall the formula for \( P_3 \) in terms of elementary symmetric sums:
\[ P_3 = S_1^3 - 3S_1S_2 + 3S_3 \]

Given \( P_3 = 36 \), \( S_1 = 6 \), and \( S_2 = 11 \), I can substitute:
\[ 36 = 6^3 - 3 \times 6 \times 11 + 3S_3 \]
\[ 36 = 216 - 198 + 3S_3 \]
\[ 36 = 18 + 3S_3 \]
\[ 3S_3 = 36 - 18 = 18 \]
\[ S_3 = 6 \]

### Step 4: Find \( P_4 \) Using Recurrence Relation
I need a recurrence relation to find \( P_4 \). For three variables, the recurrence relation is:
\[ P_k = S_1 P_{k-1} - S_2 P_{k-2} + S_3 P_{k-3} \]

I already have \( P_1, P_2, P_3 \). Let's find \( P_4 \) using \( k = 4 \):
\[ P_4 = S_1 P_3 - S_2 P_2 + S_3 P_1 \]

Substituting the known values:
\[ P_4 = 6 \times 36 - 11 \times 14 + 6 \times 6 \]
\[ P_4 = 216 - 154 + 36 \]
\[ P_4 = (216 - 154) = 62 \]
\[ 62 + 36 = 98 \]

### Verification
Just to make sure, I can think about the roots of the polynomial whose roots are \( x, y, z \). The polynomial would be:
\[ t^3 - S_1 t^2 + S_2 t - S_3 = 0 \]
\[ t^3 - 6t^2 + 11t - 6 = 0 \]

Let's find the roots:
- Try \( t = 1 \):
  \[ 1 - 6 + 11 - 6 = 0 \] → \( t = 1 \) is a root.
- Factor out \( (t - 1) \):
  \[ t^3 - 6t^2 + 11t - 6 = (t - 1)(t^2 - 5t + 6) \]
- Factor the quadratic:
  \[ t^2 - 5t + 6 = (t - 2)(t - 3) \]

So the roots are \( t = 1, 2, 3 \).

Now, compute \( P_4 \) directly:
\[ 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98 \]

This matches the result from the recurrence relation, so I can be confident.

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