X+y+z=1......(1)

Question

x^2+y^2+z^2=35........(2)
x^3+y^3+z^3=97.........(3)
solution but from 3
(x^3+y^3)=(x+y)^3-3xy(x+y)......(4)
puting 4 into 3 bck
(x+y)^3-3xy(x+y)+z^3=97.......(5)
now from 1
x+y=1-z......(6)
putin 6 into 7 we have
(1-z)^3-3xy(1-z)+z^3=97
[1-2z+z^2](1-z)-3xy(1-z)+z^3=97
1(1-2z+z^2)-z(1-2z+z^2)-3xy(1-z)+z^3=97
1-2z+z^2-z+2z^2-z^3-3xy(1-z)+z^3=97
3z^2-3z-z-3xy(1-z)=96
3z^2-4z-3xy(1-z)=96........(8)
but from 2
(x^2+y^2)=(x+y)^2-2xy.........(9)
putin 9 bck to 2
(x+y)^2-xy+z^2=35......(10)
puting 6 into 10
(1-z)^2-xy+z^2=35
[1-2z+z^2)-2xy+z^2=35
2z^2-2z-2xy=34
z^2-z-xy=17
-xy=17-z^2-z
-xy=17-z^2+z
xy=-17+z^2-z
xy=z^2-z-17
putin xy into8
3z^2-3z-z-3(z^2-z-17)(1-z)=96
3z^2-4z-3(z^2-z-17)(1-z)=96....plz help me finish it

Answer 1

Plz it is 3z not 4z

Answer 2

well, if we want integer solutions, it is clear that not all the values can be positive. However, squares are positive, and we know that

1^2+3^2+5^2 = 35

You will need to make one or two of the variables negative, so you sum to 1, but that should work for the cubes as well.

Answer 3

Steve can u please just interpret what you are saying for me please..so that i can continue

Answer 4

well, using the values 1,3,5

how can you make them add up to 1 if not all are positive?

Answer 5

Yez i got that steve thanks so much wish to be like you someday