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| 1. |
X³+y³+z³=k (here k is 42) |
| Answer» SOLUTION x=0 , y=0 and z=1 and k= 1For 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) | |