# Simultaneous equation

posted by Collins

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

1. Collins

Plz it is 3z not 4z

2. Steve

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.

3. Collins

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

4. Steve

well, using the values 1,3,5

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

5. Collins

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

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