Question

if x+y+z=1 , xy+yz+zx=-1and xyz= -1 find the value of x³+y³+z³

Answer 1

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

x3+y3+z3=(x +y + z) (x2+y2+ z2yz zx xy) + 3xyz ( Rearranging the terms)

Putting the values,

x3+y3+z3= (1) {(x2+y2+ z2- ( -1)} +3 (-1)

x3+y3+z3= (1) (x2+y2+ z2+1 ) -3 ------------------- Equation1

Now,

x2+y2+ z2

(x+y+z)2 =x2+y2+ z2+ 2 (xy+yz+zx)

(1)2 =x2+y2+ z2+2(-1)

1 =x2+y2+ z2-2

3=x2+y2+ z2----------------------------- Equation 2

Putting values of 2 on 1, we get

x3+y3+z3= (1) (3+1 ) -3

x3+y3+z3= (1) (4) -3

x3+y3+z3 = 1

Answer 2

Question-

If x + y + z = 1, xy + yz + zx = -1 and xyz= -1, find the value of x³ + y³ + z³.

Answer-

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)  

=> x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² + 2xy + 2yz + 2zx – 3xy – 3yz – 3zx)  

(Subtracting and adding 2xy + 2yz + 2zx)  

=> x³ + y³ + z³ – 3xyz = (x + y + z) {(x + y + z)² – 3(xy + yz + zx)}  

=> x³ + y³ + z³ – 3 x (-1) = 1 x {(1) 2 – 3 x (-1)}  

[Putting x + y + z = 1; xy + yz + zx = -1; xyz = -1]  

=> x³ + y³ + z³ + 3 = 4  

=> x³ + y³ + z³ = 4 – 3

x³ + y³ + z³ = 1