Question

The number of solutions of the equations \[ x + y + z = 1,\quad x^2 + y^2 + z^2 = 1,\quad x^3 + y^3 + z^3 = 1 \] is:

• A. 6
• B. 3
• C. 9
• D. 12

Hint:

In systems with symmetric sums, use Newton's identities to convert power sums to products and roots, simplifying the solution counting.

Answer 1

The Correct Option is B

We are given a symmetric system of three equations in three variables: \[ x + y + z = 1
x^2 + y^2 + z^2 = 1
x^3 + y^3 + z^3 = 1 \] Using Newton's identities, we connect power sums to elementary symmetric sums. Let: \[ S_1 = x + y + z = 1,\quad S_2 = x^2 + y^2 + z^2 = 1,\quad S_3 = x^3 + y^3 + z^3 = 1 \] Apply Newton’s identities: \[ S_1 = 1
S_2 = S_1^2 - 2(xy + yz + zx) \Rightarrow 1 = 1 - 2(xy + yz + zx) \Rightarrow xy + yz + zx = 0 \] Also, for symmetric functions: \[ \begin{align} S_3 = x^3 + y^3 + z^3 = S_1^3 - 3S_1(xy + yz + zx) + 3xyz \Rightarrow 1 = 1 - 0 + 3xyz \Rightarrow xyz = 0 \] So the roots of the polynomial with symmetric sums: \[ x + y + z = 1,\quad xy + yz + zx = 0,\quad xyz = 0 \] This means one root is 0. Let \( z = 0 \). Then: \[ x + y = 1,\quad x^2 + y^2 = 1,\quad x^3 + y^3 = 1 \] From this, solving yields three distinct solutions.