Question

x³+y³+z³=k (here k is 42)

Answer 1

Answer:

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)