Question 52
If $$x + y + z = 2, xy + yz + zx = -11$$, then the value of $$x^3 + y^3 + z^3 - 3xyz$$ is:
Solution
As we know , x^3 + y^3 +z^3- 3xyz = (x+y+z)(x^2 + y^2 +z^2 - xy -yz - zx)-------(1)
Given - (x+y+z) = 2 and xy +yz+zx= -11
(x+y+z)^2= x^2 + y^2 +z^2 +2(xy+yz+zx)
(2)^2 = x^2 + y^2 + z^2 + 2(-11)
4= x^2+ y^2 +z^2 - 22
4+22= x^2 +y^2 +z^2
26= x^2 +y^2 +z^2
putting all these values in equation (1)
x^3 + y^3 +z^3- 3xyz= 2(26-(-11))= 2(26+11) = 2(37) =74
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