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.

We are given the power sums of three real numbers $ x, y, z $:

$$
\begin{aligned}
p_1 &= x + y + z = 6, \\
p_2 &= x^2 + y^2 + z^2 = 14, \\
p_3 &= x^3 + y^3 + z^3 = 36.
\end{aligned}
$$

We are to find:

$$
p_4 = x^4 + y^4 + z^4 = ?
$$

---

### Step 1: Use Newton's identities

Let the elementary symmetric sums be:

- $ e_1 = x + y + z $
- $ e_2 = xy + yz + zx $
- $ e_3 = xyz $

Newton’s identities relate power sums $ p_k = x^k + y^k + z^k $ to elementary symmetric sums:

$$
\begin{aligned}
p_1 &= e_1, \\
p_2 &= e_1 p_1 - 2e_2, \\
p_3 &= e_1 p_2 - e_2 p_1 + 3e_3, \\
p_4 &= e_1 p_3 - e_2 p_2 + e_3 p_1.
\end{aligned}
$$

(For $ n=3 $ variables, the general formula for $ k \le 3 $ includes terms up to $ e_k $, and for $ k > 3 $, it continues with $ e_3 $ but no higher.)

We know $ p_1 = 6 $, $ p_2 = 14 $, $ p_3 = 36 $. Let's compute $ e_1, e_2, e_3 $.

- $ e_1 = p_1 = 6 $

From $ p_2 = e_1 p_1 - 2e_2 $:

$$
14 = 6 \cdot 6 - 2e_2 \Rightarrow 14 = 36 - 2e_2 \Rightarrow 2e_2 = 36 - 14 = 22 \Rightarrow e_2 = 11.
$$

From $ p_3 = e_1 p_2 - e_2 p_1 + 3e_3 $:

$$
36 = 6 \cdot 14 - 11 \cdot 6 + 3e_3 = 84 - 66 + 3e_3 = 18 + 3e_3.
$$

So:

$$
3e_3 = 36 - 18 = 18 \Rightarrow e_3 = 6.
$$

Now compute $ p_4 $ using:

$$
p_4 = e_1 p_3 - e_2 p_2 + e_3 p_1.
$$

Plug in values:

$$
p_4 = 6 \cdot 36 - 11 \cdot 14 + 6 \cdot 6 = 216 - 154 + 36 = (216 + 36) - 154 = 252 - 154 = 98.
$$

---

### ✅ Final Answer:

$$
\boxed{98}
$$