If x + y + z = 1, x^2 + y^2 + z^2 = 2, and x^3 + y^3 + z^3 = 3, what is x^5 + y^5 + z^5?

Steps to Solve

1. Identify known sums

We have the following known sums:

• $x + y + z = 1$
• $x^2 + y^2 + z^2 = 2$
• $x^3 + y^3 + z^3 = 3$

2. Use the relationship for the sums of powers

There is a known relationship between these sums.

The equation for the sum of cubes can be expressed as: $$ x^3 + y^3 + z^3 = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx) $$

3. Calculate $xy + yz + zx$ using $x^2 + y^2 + z^2$

We can derive the value of $xy + yz + zx$ from the equation: $$ x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy + yz + zx) $$

Plugging known values into the equation: $$ 2 = 1^2 - 2(xy + yz + zx) $$ This simplifies to: $$ 2 = 1 - 2(xy + yz + zx) $$ Rearranging gives: $$ 2(xy + yz + zx) = -1 $$ Thus, $$ xy + yz + zx = -\frac{1}{2} $$

4. Calculate $x^3 + y^3 + z^3$ using known sums

Using the earlier derived sums now helps us find $x^3 + y^3 + z^3$: $$ x^3 + y^3 + z^3 = (1)(2 + \frac{1}{2}) = 3 $$

5. Calculate $x^4 + y^4 + z^4$ using the known sums and previous result

We can use the recurrence relation for power sums: $$ S_n = (x+y+z) S_{n-1} - (xy + xz + yz) S_{n-2} $$

Using $S_1$, $S_2$, and $S_3$:

• $S_1 = 1$
• $S_2 = 2$
• $S_3 = 3$

Calculate $S_4$: $$ S_4 = 1 \cdot 3 - \left(-\frac{1}{2}\right) \cdot 2 $$ This becomes: $$ S_4 = 3 + 1 = 4 $$

6. Calculate $x^5 + y^5 + z^5$

Again use the recurrence for $S_5$: $$ S_5 = (x+y+z) S_4 - (xy + xz + yz) S_3 $$

Substituting known values: $$ S_5 = 1 \cdot 4 - \left(-\frac{1}{2}\right) \cdot 3 $$ This results in: $$ S_5 = 4 + \frac{3}{2} = 4 + 1.5 = 5.5 $$