1.

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

Answer»

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

Given, x + y + z = 1, xy +yz + zx = -1 and xyz = -1

To find, x2 + y2 + z2, we use the identity :

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

⇒ 1 =  x2 + y2 + z2 + (2 x -1)

⇒  x2 + y2 + z2 = 1 + 2 = 3

∴ Substituting all the values in eqn (i), we get

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

⇒ 1 = (3 + 1) = x3 + y3 + z3 +3

⇒ 4 = x3 + y3 + z3 + 3 ⇒ x3 + y3 + z3 = 1.


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