Question

Use Euclid’s division lemma to show that cube of any positive integer is of the form
9m, 9m + 1, 9m + 8.

Answer 1

Answer:

$\huge\tt\green{❥}\tt\pink{s}\tt\blue{o}\tt\pink{l}\tt\purple{u}\tt\orange{t}\tt\orange{ion}\huge\tt\purple{꧂{}}$

Let a and b be two positive integers, and a>b

a=(b×q)+r where q and r are positive integers and 

0≤r<b

Let b=3 (If 9 is multiplied by 3 a perfect cube number is obtained) 

a=3q+r where 0≤r<3

(i) if r=0,a=3q    (ii) if r=1,a=3q+1      (iii) if r=2,a=3q+2

Consider, cubes of these

Case (i) a=3q

a3=(3q)3=27q3=9(3q3)=9m           where m=3q3 and   'm' is an integer.

Case (ii) a=3q+1

a3=(3q+1)3                [(a+b)3=a3+b3+3a2b+3ab2]

      =27q3+1+27q2+9q=27q3+27q2+9q+1

      =9(3q

Answer 2

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

$a {}^{3}  = (3q) {}^{3} \\</p><p>a {}^{3}  = 27 q {}^{3} \\ </p><p> a {}^{3}  = 9 (3q {}^{3} )\\ </p><p>\sf{a {}^{3} = 9m} \\ </p><p> \sf{where \: m = 3q {}^{3} }</p><p>$

Case 2: When r = 1, the equation becomes

a = 3q + 1

$a {}^{3}  = (3q + 1) {}^{3} \\ </p><p>a {}^{3}  = (3q) {}^{3}   \\ + 1 {}^{3}  + 3 × 3q × 1(3q + 1) \\ </p><p>a {}^{3} = 27q {}^{3}  + 1 + 9q × (3q + 1) \\ </p><p>a {}^{3} = 27q {}^{3}  + 1 + 27q {}^{3} + 9q \\ </p><p>a3 = 27q3 + 27q2 + 9q + 1 \\ </p><p>a3 = 9 ( 3q3 + 3q2 + q) + 1 \\ </p><p>$

$\sf {Where \: m = ( 3q {}^{3}  + 3q {}^{2}  + q)}$

Cubing both the sides

$a {}^{3}  = (3q + 2) {}^{3} </p><p>a {}^{3}  = (3q) {}^{3}  + 2 {}^{3}  + 3 × 3q × 2 (3q + 1) </p><p>a {}^{3} = 27q {}^{3} + 8 + 54q {}^{2}  + 36q</p><p>a3 = 27q3 + 54q2 + 36q + 8</p><p>a3 = 9 (3q3 + 6q2 + 4q) + 8</p><p>a3 = 9m + 8</p><p>$

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

Hope it helps✔️