Question

4. Use division algorithm, to show that cube of any positive integer is of the form 9m,9m+1 or 9m+8?

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Answer 1

$\huge\blue{\Large{\boxed{\tt{ \red{Solution:- }}}}}$

Let,

$\red\succ$a and b where a be any positive number

where, b = 3 and r = 0,1,2 & 3

By using to Euclid’s Division Lemma,

$\tt\boxed{a = bq + r (0 ≤ r < b)}$

$\tt\boxed{a = 3q + r}$

$\underline\green{\underline{\sf{\maltese\: Case\:1\::-}}}$

(When r = 0, the equation becomes:-

a = 3q)

Cubing both the sides:-

$\purple\implies\sf{{a}^{3} = {(3q)}^{3}}$

$\purple\implies\sf{{a}^{3} = {27q}^{3}}$

$\purple\implies\sf{{a}^{3} = 9 ({3q}^{3})}$

$\purple\implies\sf{{a}^{3} = 9m}</p><p>(where,\tt{m = 3q^3})$

$\underline\green{\underline{\sf{\maltese\: Case\:2\::-}}}$

(When r = 1, the equation becomes:-

a = 3q + 1)

Cubing both sides:-

$\purple\implies\sf{{a}^{3} = (3q + 1)^{3}}$

$\purple\implies\sf{a^3 = (3q)^3+ (1)^3+ 3 × 3q × 1(3q + 1)}$

$\purple\implies\sf{a^3= 27q^3 + 1 + 9q × (3q + 1)}$

$\purple\implies\sf{a^3= 27q^3 + 1 + 27q^2 + 9q}$

$\purple\implies\sf{a^3 = 27q^3+ 27q^2 + 9q + 1}$

$\purple\implies\sf{a^3 = 9 ( 3q^3 + 3q^2 + q) + 1}$

$\purple\implies\sf{a^3= 9m + 1}$

(where,$\tt{m = ( 3q^3+ 3q^2+ q)})$

$\underline\green{\underline{\sf{\maltese\: Case\:3\::-}}}$

(When r = 2, the equation becomes:-

a = 3q + 2)

Cubing both the sides:-

$\purple\implies\sf{a^3= (3q + 2)^3}$

$\purple\implies\sf{a^3 = (3q)^3+ (2)^3 + 3 × 3q × 2 (3q + 1)}$

$\purple\implies\sf{a^3 = 27q^3 + 8 + 54q^2+ 36q}$

$\purple\implies\sf{a^3 = 27q^3 + 54q^2 + 36q + 8}$

$\purple\implies\sf{a^3 = 9 (3q^3+ 6q^2+ 4q) + 8}$

$\purple\implies\sf{a^3 = 9m + 8}$

Where, m = (3q³+ 6q²+ 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.

Identity used:-

• (a + b)³ = a³ + b³ + 3ab (a + b)

Hope it helps :)

Answer 2

Answer:

Answer:

Let us consider a and b where a be any positive number and b is equal to 3.

According to Euclid’s Division Lemma

a = bq + r

where r is greater than or equal to zero and less than b (0 ≤ r < b)

a = 3q + r

so r is an integer greater than or equal to 0 and less than 3.

Hence r can be either 0, 1 or 2.

Case 1: When r = 0, the equation becomes

a = 3q

Cubing both the sides

a3 = (3q)3

a3 = 27 q3

a3 = 9 (3q3)

a3 = 9m

where m = 3q3

Case 2: When r = 1, the equation becomes

a = 3q + 1

Cubing both the sides

a3 = (3q + 1)3

a3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1)

a3 = 27q3 + 1 + 9q × (3q + 1)

a3 = 27q3 + 1 + 27q2 + 9q

a3 = 27q3 + 27q2 + 9q + 1

a3 = 9 ( 3q3 + 3q2 + q) + 1

a3 = 9m + 1

Where m = ( 3q3 + 3q2 + q)

Case 3: When r = 2, the equation becomes

a = 3q + 2

Cubing both the sides

a3 = (3q + 2)3

a3 = (3q)3 + 23 + 3 × 3q × 2 (3q + 1)

a3 = 27q3 + 8 + 54q2 + 36q

a3 = 27q3 + 54q2 + 36q + 8

a3 = 9 (3q3 + 6q2 + 4q) + 8

a3 = 9m + 8

Where m = (3q3 + 6q2 + 4q)therefore a can be any of the form 9m or 9m + 1 or, 9m + 8.

Step-by-step explanation: