InterviewSolution
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InterviewSolution
| 1. |
Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100 |
| Answer» <html><body><p>We understand x , y and z as integers belonging to IZ => negative and positive Natural integers.x3+y3+z3=kx3+y3+z3=k with k is <a href="https://interviewquestions.tuteehub.com/tag/integer-11854" style="font-weight:bold;" target="_blank" title="Click to know more about INTEGER">INTEGER</a> from 1 to 100solution x=0 , y=0 and z=1 and k= <a href="https://interviewquestions.tuteehub.com/tag/1for-282962" style="font-weight:bold;" target="_blank" title="Click to know more about 1FOR">1FOR</a> 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,-<a href="https://interviewquestions.tuteehub.com/tag/8-336412" style="font-weight:bold;" target="_blank" title="Click to know more about 8">8</a>,-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)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)2.z−3(x+y).z2=k=>(x+y+z)3−3(x+y)z[(x+y)−3z]=k=>(x+y+z)3−3(x+y)z[(x+y)−3z]=klety=αy=αand z=βz=β=>x3=−α3−β3+k=>x3=−α3−β3+kFor 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= <a href="https://interviewquestions.tuteehub.com/tag/28-299271" style="font-weight:bold;" target="_blank" title="Click to know more about 28">28</a> => (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 = <a href="https://interviewquestions.tuteehub.com/tag/35-308504" style="font-weight:bold;" target="_blank" title="Click to know more about 35">35</a> 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 =36we have also solution : x=1,y=2andz=3=>x=1,y=2andz=3=>13+23+33=1+8+</p></body></html> | |