show that cube of a positive integer is of the form 6 m + r where r = 0, 1, 2.......5
Answers
So according to Euclid's division lemma,
a = bq + r
Formula for ( a + b)³ = (a)³ + 3a²b + 3ab²+ (b)³
1.) r = 0
a = 6q + 0
Cubing both the sides,
(a)³ = ( 6q)³
taking q³ = m
a³ = 6m+0
2.) r = 1
a = 6q+ 1
Cubing both the sides,
a³= (6q+1)³
a³= 6q³ + 3×6q²×1 + 3×6q×1² + 1³
a³= 216q³ + 108q² + 18q + 1
a³= 6q(36q² + 18q + 3) + 1
taking q(6q + 18q +3) = m
a³= 6m + 1
3.) r = 2
a = 6q+ 2
Cubing both the sides,
a³= (6q+2)³
a³= 6q³+ 3×6q²×2 + 3×6q×2² + 2³
a³= 216q³ + 216q² + 72q + 8
a³= 216q³ + 216q² + 72q + 6 + 2
a³= 6(36q³ + 36q² + 12q + 1) + 2
taking 36q³ + 36q² + 12q + 1) = m
a³= 6m + 2
4.) r = 3
a = 6q+3
Cubing both the sides,
a³.= (6q+3)³
a³= 6q³+ 3×6q²×3 + 3×6q×3² + 3³
a³= 216q³ + 324q² + 162q + 9
a³= 216q³ + 324q² + 162q + 6 + 3
a³= 6(36q³ + 54q² + 27q + 1) + 3
taking 36q³ + 54q² + 27q + 1) = m
a³= 6m+3
5.) r = 4
a = 6q + 4
Cubing both the sides,
a³ = (6q+ 4)³
a³= 6q³+ 3×6q²×4 + 3×6q×4² + 4³
a³= 216q³ + 432q² + 288q + 64
a³= 216q³ + 432q² + 288q + 60 + 4
a³= 6(36q³+ 72q² + 48q + 10) + 4
taking 36q³ + 72q² + 48q + 10) = m
a³= 6m + 4
6.) r = 5
a = 6q + 5
Cubing both the sides,
a³= (6q+ 5)³
a³ = 6q³+ 3×6q²×5 + 3×6q×5² + 5³
a³= 216q³ + 540q² + 450q + 125
a³= 216q³ + 540q² + 450q + 120 + 5
a³= 6(36q³ + 90q² + 75q + 20) + 5
taking 36q³ + 90q² + 75q + 20) = m
a³= 6m + 5
HOPE THIS HELPS YOU...!!!