To solve the problem, we use Newton's identities relating power sums to elementary symmetric sums. Let's denote:

Step 1: Define Symmetric Sums

Let (S_1 = x+y+z), (S_2 = xy+yz+zx), (S_3 = xyz), and (P_n = x^n + y^n + z^n).

Given:
(P_1 = 1), (P_2 = 2), (P_3 = 3).

Step 2: Compute (S_2)

[P_2 = S_1^2 - 2S_2]
[2 = 1^2 - 2S_2 \implies 2 = 1 - 2S_2 \implies S_2 = -\frac{1}{2}]

Step 3: Compute (S_3)

[P_3 - 3S_3 = S_1(P_2 - S_2)]
[3 - 3S_3 = 1\left(2 - \left(-\frac{1}{2}\right)\right) = \frac{5}{2}]
[3S_3 = 3 - \frac{5}{2} = \frac{1}{2} \implies S_3 = \frac{1}{6}]

Step 4: Compute (P_4)

Using Newton's identity:
[P_4 = S_1P_3 - S_2P_2 + S_3P_1]
[P_4 = 1 \cdot 3 - \left(-\frac{1}{2}\right) \cdot 2 + \frac{1}{6} \cdot 1]
[P_4 = 3 + 1 + \frac{1}{6} = \frac{25}{6}]

Answer: (\boxed{\dfrac{25}{6}})

作者声明:本文包含人工智能生成内容。