Question

x+y+z=1, xy+ya+zx= -1and xyz=-1.find the value of x^3+6^3+z^3

Answer 1

x³ + y³ + z³ - 3zyz = (x + y + z) (x² + y² + z² - (xy + ya + zx)

Here we have no value of x² + y² + z² so we will find that first

(x + y + z)² = x² + y² + z² + 2 (xy + ya + zx)

1² = x² + y² + z² + 2 x -1

x² + y² + z² = 1 + 2 = 3

Now x³ + y³ + z³ - 3zyz = (x + y + z) (x² + y² + z² - (xy + ya + zx))

x³ + y³ + z³ - 3 x -1 = 1 ( 3 - -1)

x³ + y³ + z³ + 3 = 4

x³ + y³ + z³ = 4 - 3 = 1