If x + y + z = 1 , xy + yz + zx = -1 , and xyz = -1 , then find the value of (x^3 + y^3 + z^3) .

Question

Accepted Answer

We know the standard algebraic identity:

x^{3} + y^{3} + z^{3} - 3 x yz = (x + y + z) (x^{2} + y^{2} + z^{2} - (x y + yz + z x))

**Step 1: Find the value of

x^{2} + y^{2} + z^{2}

**
Using the identity:

(x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + yz + z x)

Substitute the given values:

(1)^{2} = x^{2} + y^{2} + z^{2} + 2 (- 1)

1 = x^{2} + y^{2} + z^{2} - 2

x^{2} + y^{2} + z^{2} = 1 + 2 = 3

**Step 2: Find the value of

x^{3} + y^{3} + z^{3}

**
Substitute the values into the main identity:

x^{3} + y^{3} + z^{3} - 3 (- 1) = (1) (3 - (- 1))

x^{3} + y^{3} + z^{3} + 3 = 1 \times (3 + 1)

x^{3} + y^{3} + z^{3} + 3 = 4

x^{3} + y^{3} + z^{3} = 4 - 3 = 1

If $x + y + z = 1$ , $x y + yz + z x = - 1$ , and $x yz = - 1$ , then find the value of $(x^{3} + y^{3} + z^{3})$ .