I don't think you will find any solutions where all three are nonzero. So, play around with
x+y = 2
x^2+y^2 = 3
x^3+y^3 = 5
Then you can use the fact that
x^3+y^3 = (x+y)(x^2-xy+y^2) = 2(3-xy) = 5
You should get somewhere with that.
X+y+z=2
x^2+y^2+z^2=3
x^3+y^3+z^3=5
find x,y,z.....??
honestly i having being battle with that question for ova a month i need help and a cerious explanation
1 answer