Question

show that cube of a positive integer is of the form 6 m + r where r = 0, 1, 2.......5​

Answer 1

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...!!!