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.

Let $S_k = x^k + y^k + z^k$ denote the power sums of the variables $x, y, z$. We are given:
$$ S_1 = 6 $$
$$ S_2 = 14 $$
$$ S_3 = 36 $$

We want to find $S_4 = x^4 + y^4 + z^4$.

Let $e_1, e_2, e_3$ be the elementary symmetric polynomials in $x, y, z$:
$$ e_1 = x + y + z $$
$$ e_2 = xy + yz + zx $$
$$ e_3 = xyz $$

We can use Newton's sums to relate the power sums $S_k$ to the elementary symmetric polynomials $e_k$. The relations are:
1.  $S_1 - e_1 = 0$
2.  $S_2 - e_1 S_1 + 2e_2 = 0$
3.  $S_3 - e_1 S_2 + e_2 S_1 - 3e_3 = 0$
4.  $S_4 - e_1 S_3 + e_2 S_2 - e_3 S_1 = 0$

**Step 1: Find $e_1$.**
From the first relation:
$$ e_1 = S_1 = 6 $$

**Step 2: Find $e_2$.**
Substitute $S_1$ and $S_2$ into the second relation:
$$ 14 - (6)(6) + 2e_2 = 0 $$
$$ 14 - 36 + 2e_2 = 0 $$
$$ -22 + 2e_2 = 0 $$
$$ 2e_2 = 22 $$
$$ e_2 = 11 $$

**Step 3: Find $e_3$.**
Substitute $S_1, S_2, S_3, e_1, e_2$ into the third relation:
$$ 36 - (6)(14) + (11)(6) - 3e_3 = 0 $$
$$ 36 - 84 + 66 - 3e_3 = 0 $$
$$ (36 + 66) - 84 - 3e_3 = 0 $$
$$ 102 - 84 - 3e_3 = 0 $$
$$ 18 - 3e_3 = 0 $$
$$ 3e_3 = 18 $$
$$ e_3 = 6 $$

**Step 4: Find $S_4$.**
Substitute $S_1, S_2, S_3, e_1, e_2, e_3$ into the fourth relation:
$$ S_4 - (6)(36) + (11)(14) - (6)(6) = 0 $$
$$ S_4 - 216 + 154 - 36 = 0 $$
$$ S_4 - 252 + 154 = 0 $$
$$ S_4 - 98 = 0 $$
$$ S_4 = 98 $$

**Alternative Method (Finding the roots):**
The numbers $x, y, z$ are the roots of the cubic polynomial $P(t) = t^3 - e_1 t^2 + e_2 t - e_3$.
Substituting the values we found:
$$ P(t) = t^3 - 6t^2 + 11t - 6 $$
We can factor this polynomial. Testing integer roots, we find $P(1) = 1 - 6 + 11 - 6 = 0$.
Dividing by $(t-1)$, we get $t^2 - 5t + 6$, which factors into $(t-2)(t-3)$.
Thus, the roots are $x=1, y=2, z=3$ (in some order).
We can verify the given sums:
$1+2+3 = 6$
$1^2+2^2+3^2 = 1+4+9 = 14$
$1^3+2^3+3^3 = 1+8+27 = 36$
Now, calculate $x^4 + y^4 + z^4$:
$$ 1^4 + 2^4 + 3^4 = 1 + 16 + 81 = 98 $$

Both methods yield the same result.

\boxed{98}