Question

give the correct answer...?​

Answer 1

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.

Answer 2

Question:-

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

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Solution:-

Let x be any positive integer and y = 3.

By Euclid’s division algorithm, then,

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Therefore, putting the value of r, we get,

x = 3q

or

x = 3q + 1

or

x = 3q + 2

Now, by taking the cube of all the three above expressions, we get,

Case (i):

When r = 0, then,

$x^2\:=\: (3q)^3 \:= \:27q^3= 9(3q^3)= 9m;$

where m = $3q^3$

Case (ii):

When r = 1, then,

$x^3 \:= \:(3q+1)^3$

$= (3q)^3 +1^3+3×3q×1(3q+1)$

$= 27q^3+1+27q^2+9q$

Taking 9 as common factor, we get,

$x^3 = 9(3q^3+3q^2+q)+1$

Putting = m, we get,

Putting$(3q^3+3q^2+q)$= m, we get ,

$x^3$= 9m+1

Case (iii):

When r = 2, then,

$x^3 = (3q+2)^3$

$= (3q)^3+2^3+3×3q×2(3q+2)$

$= 27q^3+54q^2+36q+8$

Taking 9 as common factor, we get,

$x^3=9(3q^3+6q^2+4q)+8$

Putting $(3q^3+6q^2+4q)$= m, we get ,

$x^3$= 9m+8

Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

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Hope it helps you!☺️.