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
Plz it is 3z not 4z
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.
Steve can u please just interpret what you are saying for me please..so that i can continue
well, using the values 1,3,5
how can you make them add up to 1 if not all are positive?
Yez i got that steve thanks so much wish to be like you someday
To solve the equation, let's continue simplifying step by step:
1. Start with equation (8): 3z^2 - 4z - 3(z^2 - z - 17)(1 - z) = 96.
2. Expand the expression (z^2 - z - 17)(1 - z) using distributive property: 3z^2 - 4z - 3(z^3 - z^2 - 17z + z^2 + z + 17) = 96.
3. Simplify the terms inside parentheses: 3z^2 - 4z - 3(z^3 - 17z + 17) = 96.
4. Distribute the -3 to each term inside parentheses: 3z^2 - 4z - 3z^3 + 51z - 51 = 96.
5. Combine like terms: -3z^3 + 3z^2 + 47z - 147 = 0.
6. Move all terms to one side to obtain a cubic equation: -3z^3 + 3z^2 + 47z - 243 = 0.
At this point, you can solve this cubic equation using numerical methods or calculators capable of solving cubic equations.
Alternatively, if you have any specific values for x, y, or z, you can substitute them back into the original equations (1), (2), and (3) to verify if they satisfy the equations.