Question

show that cube of any integer can be expressed in one the form 9k,9k+1,9k+8​

Answer 1

$take \ \\ any \: integer \: can \: be \: express \\ n = 3m + 2 \ \: or \: 3m + 1 \: \: and \: \: 3m \\ {n}^{3} = {(3m + 2)}^{3} \\ \: \: \: = 27 {m}^{3} + 8 + 8m(3m + 2) \\ \: \: \: = 9(3 {m}^{3} + 6 {m}^{2} + 4m) + 8 \\ \: \: = 9k + 8 \: \: where \: \\ \: \: \: \: \: \: \: \: k = (3 {m}^{3} + 6 {m}^{2} + 4m ) \\ similarly \: you \: can \: do \: all$

Answer 2

Let the a be any positive integer when divided by 9 gives "q" as quoteint and "r" as remainder.

Then,by using Euclid division lemma

$a = bq + r \: (0 < = r < b)$

$a = 9q + r \: (r = 0 .1.2.3....)$

Case :- 1 ,when r = 0

$a = 9q$

Cubing on both side,

${a}^{3} = {(9q)}^{3}$

${a}^{3} = 729 {q}^{3}$

$let \: 9 {q}^{3} = k \: for \: some \: integer \: k$

${a}^{3} = 9k$

Case 2 ,when r = 1

$a = 9q + 1$

Cubing on both sides

${a}^{3} = {(9q + 1)}^{3}$

${a}^{3} = 729 {q}^{3} + 243q + 27q + 1$

${a}^{3} = 9(81 {q}^{3} + 27 {q}^{2} + 3q) + 1$

$let(81 {q}^{3} + 27 {q}^{2} + 3q) = k$

for some integer ,k

${a}^{3} = 9k + 1$

Case 3, when r=2,

$a = 9q + 2$

Cubing on both sides ,

${a}^{3} = {(9q + 2)}^{3}$

${a}^{3} = 729 {q}^{3} + 486 {q}^{2} + 108q + 8$

${a}^{3} = 9(81 {q}^{3} + 53 {q}^{2} + 21q) + 8$

$let(81 {q}^{3} + 53 {q}^{2} + 21q) = k$

for some integer k,

${a}^{3} = 9k + 8$

hence,the cube of every positive integer can be expressed in the form of 9k , 9k+1, 9k+8 for some integer k.