Happy Thanksgiving/Diwali!! I thank you all for your help!! Take these points as a motivation to help :) Also if you want solve k : x³ + y³ + z³ = k This is a little hard.
Answers
Answer:
We understand x , y and z as integers belonging to IZ => negative and positive Natural integers.
x3+y3+z3=k with k is integer from 1 to 100
solution x=0 , y=0 and z=1 and k= 1
For K= 1 , we have the following solutions (x,y,x) = (1,0,0) ; or (0,1,0) ; or (0,0,1) ,
For k =1 also (9,-8,-6) or (9,-6,-8) or (-8,-6,9) or (-8,9,-6) or (-6,-8,9) or (-6,9,8)
And (-1,1,1) or (1,-1,1)
=>(x+y)3−3x2−3xy2+z3=k
=>(x+y+z)3−3(x+y)2.z−3(x+y).z2=k
=>(x+y+z)3−3(x+y)z[(x+y)−3z]=k
let y=α and z=β
=>x3=−α3−β3+k
For k= 2 we have (x,y,z) = (1,1,0) or (1,0,1) or (0,1,1)
Also for (x,y,z) = (7,-6,-5) or (7,-6,-5) or (-6,-5,7) or (-6,7,-5) or (-5,-6,7) or (-5,7,-6)
For k= 3 we have 1 solution : (x,y,z) = (1,1,1)
For k= 10 , we have the solutions (x,y,z) = (1,1,2) or (1,2,1) or (2,1,1)
For k= 9 we have the solutions (x,y,z) = (1,0,2) or (1,2,0) or (0,1,2) or (0,2,1) or (2,0,1) or (2,1,0)
For k= 8 we have (x,y,z) = ( 0,0,2) or (2,0,0) or (0,2,0)
For k= 17 => (x,y,z) = (1,2,2) or (2,1,2) or ( 2,2,1)
For k = 24 we have (x,y,z) = (2,2,2)
For k= 27 => (x,y,z) = (0,0,3) or (3,0,0) or (0,3,0)
for k= 28 => (x,y,z) = (1,0,3) or (1,3,0) or (1,3,0) or (1,0,3) or (3,0,1) or (3,1,0)
For k=29 => (x,y,z) = (1,1,3) or (1,3,1) or (3,1,1)
For k = 35 we have (x,y,z) = (0,2,3) or (0,3,2) or (3,0,2) or (3,2,0) or 2,0,3) or (2,3,0)
For k =36
we have also solution : x=1,y=2andz=3=>
13+23+33=1+8+27=36 with k= 36 , we have the following
we Have : (x, y,z) = (1, 2, 3) ; (3,2,1); (1,3,2) ; (2,1,3) ; (2,3,1), and (3,1,2)
For k= 43 we have (x,y,z) = (2,2,3) or (2,3,2) or (3,2,2)
For k = 44 we have ( 8,-7,-5) or (8,-5,-7) or (-5,-7,8) or ( -5,8,-7) or (-7,-5,8) or (-7,8,-5)
For k =54 => (x,y,z) = (13,-11,-7) ,
for k = 55 => (x,y,z) = (1,3,3) or (3,1,3) or (3,1,1)
and (x,y,z) = (10,-9,-6) or (10,-6,-9) or ( -6,10,-9) or (-6,-9,10) or (-9,10,-6) or (-9,-6,10)
For k = 62 => (x,y,z) = (3,3,2) or (2,3,3) or (3,2,3)
For k =64 => (x,y,z) = (0,0,4) or (0,4,0) or (4,0,0)
For k= 65 => (x,y,z) = (1,0,4) or (1,4,0) or (0,1,4) or (0,4,1) or (4,1,0) or (4,0,1)
For k= 66 => (x,y,z) = (1,1,4) or (1,4,1) or (4,1,1)
For k = 73 => (x,y,z) = (1,2,4) or (1,4,2) or (2,1,4) or (2,4,1) or (4,1,2) or (4,2,1)
For k= 80=> (x,y,z)= (2,2,4) or (2,4,2) or (4,2,2)
For k = 81 => (x,y,z) = (3,3,3)
For k = 90 => (x,y,z) = (11,-9,-6) or (11,-6,-9) or (-9,11,-6) or (-9,-6,11) or (-6,-9,11) or (-6,11,-9)
k = 99 => (x,y,z) = (4,3,2) or (4,2,3) or (2,3,4) or (2,4,3) or ( 3,2,4 ) or (3,4,2)
(x,y,z) = (5,-3,1) or (5,1,-3) or (-3,5,1) or (-3,1,5) or (1,-3,5) or (1,5,-3)
=> 5^3 + (-3)^3 +1 = 125 -27 +1 = 99 => for k = 99
For K = 92
6^3 + (-5)^3 +1 = 216 -125 +1 = 92
8^3 +(-7)^3
More solution could be extracted.
Hope it helps :)
P.S- Phew!!! I'm tired!!!
Answer:
Processes or Throughputs are the actions within the farm that allow the inputs to turn into outputs. Processes could include things such as milking, harvesting and shearing. Outputs can be negative or positive, although they are usually the latter. Negative outputs include waste products and soil erosion.