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: Find the Sum of Products \( xy + yz + zx \)
I recall that for any three numbers, the square of the sum can be expanded as:
\[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \]

Given \( x + y + z = 6 \) and \( x^2 + y^2 + z^2 = 14 \), I can plug these into the equation:
\[ 6^2 = 14 + 2(xy + yz + zx) \]
\[ 36 = 14 + 2(xy + yz + zx) \]
\[ 36 - 14 = 2(xy + yz + zx) \]
\[ 22 = 2(xy + yz + zx) \]
\[ xy + yz + zx = 11 \]

### Step 2: Find the Product \( xyz \)
Now, I need to find \( xyz \). I recall the formula for the sum of cubes:
\[ x^3 + y^3 + z^3 = (x + y + z)^3 - 3(x + y + z)(xy + yz + zx) + 3xyz \]

Given \( x^3 + y^3 + z^3 = 36 \), \( x + y + z = 6 \), and \( xy + yz + zx = 11 \), I can substitute these values in:
\[ 36 = 6^3 - 3 \times 6 \times 11 + 3xyz \]
\[ 36 = 216 - 198 + 3xyz \]
\[ 36 = 18 + 3xyz \]
\[ 36 - 18 = 3xyz \]
\[ 18 = 3xyz \]
\[ xyz = 6 \]

### Step 3: Find the Sum of Fourth Powers \( x^4 + y^4 + z^4 \)
This is a bit more complex. I recall that for any three numbers, the sum of fourth powers can be related to the previous sums. However, I'm not sure of a direct formula, so I might need to think about Newton's identities or recursive relations.

But perhaps I can use the following approach:
First, consider the square of the sum of squares:
\[ (x^2 + y^2 + z^2)^2 = x^4 + y^4 + z^4 + 2(x^2y^2 + y^2z^2 + z^2x^2) \]

I already know \( x^2 + y^2 + z^2 = 14 \), so:
\[ 14^2 = x^4 + y^4 + z^4 + 2(x^2y^2 + y^2z^2 + z^2x^2) \]
\[ 196 = x^4 + y^4 + z^4 + 2(x^2y^2 + y^2z^2 + z^2x^2) \]

Now, I need to find \( x^2y^2 + y^2z^2 + z^2x^2 \).

### Step 4: Find \( x^2y^2 + y^2z^2 + z^2x^2 \)
I recall that:
\[ (xy + yz + zx)^2 = x^2y^2 + y^2z^2 + z^2x^2 + 2xyz(x + y + z) \]

I already have \( xy + yz + zx = 11 \) and \( xyz = 6 \), and \( x + y + z = 6 \). So:
\[ 11^2 = x^2y^2 + y^2z^2 + z^2x^2 + 2 \times 6 \times 6 \]
\[ 121 = x^2y^2 + y^2z^2 + z^2x^2 + 72 \]
\[ 121 - 72 = x^2y^2 + y^2z^2 + z^2x^2 \]
\[ x^2y^2 + y^2z^2 + z^2x^2 = 49 \]

### Step 5: Substitute Back to Find \( x^4 + y^4 + z^4 \)
Now, recall from Step 3:
\[ 196 = x^4 + y^4 + z^4 + 2 \times 49 \]
\[ 196 = x^4 + y^4 + z^4 + 98 \]
\[ 196 - 98 = x^4 + y^4 + z^4 \]
\[ x^4 + y^4 + z^4 = 98 \]

### Verification
Just to make sure, I can think about the roots of the polynomial whose roots are \( x, y, z \). The elementary symmetric sums are:
1. \( S_1 = x + y + z = 6 \)
2. \( S_2 = xy + yz + zx = 11 \)
3. \( S_3 = xyz = 6 \)

The polynomial is:
\[ t^3 - S_1 t^2 + S_2 t - S_3 = 0 \]
\[ t^3 - 6t^2 + 11t - 6 = 0 \]

Let's find the roots:
Trying \( t = 1 \):
\[ 1 - 6 + 11 - 6 = 0 \] → \( t = 1 \) is a root.

Now, perform polynomial division or factor it:
\[ (t - 1)(t^2 - 5t + 6) = 0 \]

Now, factor the quadratic:
\[ t^2 - 5t + 6 = (t - 2)(t - 3) \]

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

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

This matches our previous result, so I can be confident that the answer is correct.

### Final Answer
\boxed{98}