Let ‘a’ be positive integer. Then from Euclidean lemma a = bq + r;

now consider b = 9 then 0 ≤ r < 9, it means remainder will be 0, or 1, 2, 3, 4, 5, 6, 7, or 8

So a = bq + r

⇒ a = 9q + r (for b = 9)

now cube of a = a^{3} + (9q + r)^{3}

= (9q)^{3} + 3.(9q) ^{3}r + 3. 9q.r + r ^{3}

= 9^{3}q^{3} + 3.9^{2} (q^{2}r) + 3.9(q.r) + r ^{3}

= 9[9^{2}.q^{3 }+ 3.9.q^{2} r + 3.q.r] + r^{3}

a^{3} = 9m + r^{3} (where ‘m’ = 9^{2} q^{3} + 3.9.q2r + 3.q.r)

if r = 0 ⇒ r^{3} = 0 then a^{3} = 9m + 0 = 9m

and for r = 1 ⇒ r^{3} = l^{3} then a^{3} = 9m + 1

and for r = 2 ⇒ r^{3} = 2^{3} then a^{3} = 9m + 8

for r = 3 ⇒ r^{3} , = 3^{3} ⇒ a^{3} = 9m + 27 = 9(m) where m = (9m +3)

for r = 4 ⇒ r^{3} = 4^{3} ⇒ a^{3} = 9m + 64 = (9m + 63) + 1 = 9m + 1

for r = 5 ⇒ r^{3} = 125 ⇒ a^{3} = 9m + 125 = (9m + 117) + 8 = 9m + 8

for r = 6 ⇒ r^{3}— 216 ⇒ a^{3} = 9m + 216 = 9m + 9(24) = 9m

for r = 7 ⇒ r^{3} = 243

⇒ a^{3} = 9m + 9(27) = 9m

for r = 8 ⇒ r^{3} = 512

⇒ a^{3} = 9m + 9(56) + 8 = 9m + 8

So from the above it is clear that a^{3} is either in the form of 9m or 9m + 1 or 9m + 8.

**Hence proved.**