Question

(5) Use Euclid's division lemma to show that the cube of any positive integer is of
the form 9m, 9m+1 or 9m+8.

Answer 1

Consider an arbitrary positive integer $\red{\sf a}$

Then, a is of the form 3q, (3q+1), or (3q+2), where q is some integer.

Now,

$\red{\sf a = 3q}$ and taking cube on both side we get,

a³ = (3q)³

   = 27q³

   = 9(3q³)  

   = 9m  ______(i)

[Taking $\green{(3q^{3}) = m}$, where m is some integer]

Now,

Consider $\red{\sf a = (3q+1)}$

a³ = (3q+1)³

   = 27q³ + 27q² + 9q +1   [ (a+b)³ = a³ + 3a²b + 3ab² + b³ ]  

   = 9(3q³ + 3q² + q) + 1  

   = 9m + 1  _______(ii)

[ ∵ $\green{(3q^{3} + 3q^{2} + q) = m}$, where m is some integer ]

Now,

Consider $\red{\sf a = (3q+2)}$)

a³ = (3q+2)³

   = 27q³ + 54q² + 36q + 8

   = 9(3q³ + 6q² + 4q) + 8

   = 9m + 8  _____(iii)

[ Taking $\green{(3q^{3} + 6q^{3} + 4q) = m}$, where m is some integer ]

So,

From eq. (i), (ii), (iii) we get,

a³ = 9m, (9m+1) and (9m+8)

∴ The cube of any positive integer is either of the form 9m, (9m+1), and (9m+8)

Hence Proved