We are tasked with solving for \( x^5 + y^5 + z^5 \) given the following system of equations:
\[
\begin{aligned}
1. & \quad x + y + z = 1, \\
2. & \quad x^2 + y^2 + z^2 = 2, \\
3. & \quad x^3 + y^3 + z^3 = 3.
\end{aligned}
\]
Step 1: Use the identity for the sum of squares
The sum of squares can be expressed in terms of the sum and the product of the variables:
\[
x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx).
\]
Substitute the given values:
\[
2 = 1^2 - 2(xy + yz + zx).
\]
Simplify:
\[
2 = 1 - 2(xy + yz + zx) \implies 2(xy + yz + zx) = -1 \implies xy + yz + zx = -\frac{1}{2}.
\]
Step 2: Use the identity for the sum of cubes
The sum of cubes can be expressed as:
\[
x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) + 3xyz.
\]
Substitute the known values:
\[
3 = 1 \left( 2 - \left( -\frac{1}{2} \right) \right) + 3xyz.
\]
Simplify:
\[
3 = 1 \left( 2 + \frac{1}{2} \right) + 3xyz = 1 \cdot \frac{5}{2} + 3xyz = \frac{5}{2} + 3xyz.
\]
Solve for \( xyz \):
\[
3 = \frac{5}{2} + 3xyz \implies 3xyz = 3 - \frac{5}{2} = \frac{6}{2} - \frac{5}{2} = \frac{1}{2} \implies xyz = \frac{1}{6}.
\]
Step 3: Use Newton's Sums to find \( x^5 + y^5 + z^5 \)
Newton's Sums provide a recursive way to compute symmetric power sums. Let \( P_k = x^k + y^k + z^k \). The recurrence relation is:
\[
P_k = -(a_{n-1})P_{k-1} - (a_{n-2})P_{k-2} - \cdots - (a_0)P_{k-n},
\]
where \( a_i \) are the coefficients of the polynomial whose roots are \( x, y, z \). The polynomial is:
\[
t^3 - (x+y+z)t^2 + (xy+yz+zx)t - xyz = 0,
\]
which becomes:
\[
t^3 - t^2 - \frac{1}{2}t - \frac{1}{6} = 0.
\]
Thus, the coefficients are \( a_2 = -1 \), \( a_1 = -\frac{1}{2} \), and \( a_0 = -\frac{1}{6} \).
Using Newton's Sums:
\[
P_1 = x + y + z = 1,
\]
\[
P_2 = x^2 + y^2 + z^2 = 2,
\]
\[
P_3 = x^3 + y^3 + z^3 = 3.
\]
For \( k = 4 \):
\[
P_4 = -(a_2)P_3 - (a_1)P_2 - (a_0)P_1 = -(-1)(3) - \left( -\frac{1}{2} \right)(2) - \left( -\frac{1}{6} \right)(1) = 3 + 1 + \frac{1}{6} = 4 + \frac{1}{6} = \frac{24}{6} + \frac{1}{6} = \frac{25}{6}.
\]
For \( k = 5 \):
\[
P_5 = -(a_2)P_4 - (a_1)P_3 - (a_0)P_2 = -(-1)\left( \frac{25}{6} \right) - \left( -\frac{1}{2} \right)(3) - \left( -\frac{1}{6} \right)(2) = \frac{25}{6} + \frac{3}{2} + \frac{2}{6}.
\]
Convert \( \frac{3}{2} \) to a fraction with denominator 6:
\[
\frac{3}{2} = \frac{9}{6}.
\]
Thus:
\[
P_5 = \frac{25}{6} + \frac{9}{6} + \frac{2}{6} = \frac{36}{6} = 6.
\]
Final Answer:
\[
\boxed{6}
\]
Parent Tip: Review the logic above to help your child master the concept of hard math.
