Question

x³+y³+z³=k,find out k in the following equation where k should be in between 1 and 100​

Answer 1

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 you✌