Question

if X + Y + Z = 5, X Y + Y Z +ZX =1 and X Y Z =-1 find the value of x cube + y cube + Z cube.

Answer 1

X³+Y³+Z³-3XYZ= (X+Y+Z)(X²+Y²+Z²-XY-YZ-ZX)

REARRANGING:

X³+Y³+Z³=(X+Y+Z)(X²+Y²+Z²-XY-YZ-ZX)+3XYZ

X³+Y³+Z³=(5)(X²+Y²+Z²-(1))+3(-1)

= 5(X²+Y²+Z² -1)-3

= (X²+Y²+Z² -1)2 ------------------------------EQUATION 1

(X+Y+Z)² = X²+Y²+Z²+2(XY+YZ+ZX)

25= X²+Y²+Z² +2*1

X²+Y²+Z²= 25-2=23

REPLACING VALUES IN EQUATION 1,

(23-1)2

= 22*2

THEREFORE, X³+Y³+Z³=44