If x+y+z=1, x y+y z+z x=x y z=-4, then what is the value of ≤ft(x^3+y…

Given,

x + y + z = 1,

xy + yz + zx = xyz = -4

We know that,

x³ + y³ + z³ – 3xyz = (x + y + z) [(x + y + z)² – 3(xy + yz + zx)]

According to the question

x³ + y³ + z³  – 3xyz = (x + y + z) [(x + y + z)² – 3(xy + yz + zx)]

x³ + y³ + z³  – 3(-4) = (1) [(1)² – 3(-4)]

x³ + y³ + z³ + 12 = (1 + 12)

x³ + y³ + z³  + 12 = 13

x³ + y³ + z³ = (13 – 12)

 (x³ + y³ + z³ ) = 1