Question

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

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

Step-by-step explanation:

Solution -

Let a Be Any Positive Integer , b=3

By Euclid's Division Lemma

i.e.

a = bq + r ....(1)

where

$0 \leqslant r < b$

But We Know That b = 3

So Remaining Values For r Are -

0,1,2

Putting The Value Of r In Equation (1)

If r = 0

Then,

a = 3q+0

a = 3q

Cubing Both Side

a³ = (3q)³

a³ = 27q³

a³ = 9 (3q³) [ Common 9 From 27).

a³ = 9m (m is Some Integer so m=3q³)

$\bold{\boxed{a3=9m}}$

If r = 1

then

a = 3q+1

Cubing Both Side

a³ = (3q+1)³

a³ = 27q³ + (1)³ + 3×(3q)²×1 + 3×(1)²×3q

a³ = 27q³ + 1 + 27q² + 9q

These Can Be Rewritten As

a³ = 27q³ + 27q² + 9q + 1

a³ = 9(3q³ + 3q² + q ) + 1 ( Common 9 From These eq )

a³ = 9m + 1 ( m is Some Integer So m= 3q³+3q²+q)

$\bold{\boxed{a3 = 9m +1}}$

If r = 2

a = 3q+ 2

Cubing Both Side

a³ = (3q+ 2)³

a³ = 27q³ + 8 + 3×(3q)²×2 + 3×(2)²×3q

a³ = 27q³ + 8 +54q² + 36q

These Can Be Rewritten As

a³ = 27q³ + 54q² + 36q + 8

a³ = 9(3q³ + 6q² + 4q) + 8 (Common 9 From These eq)

a³ = 9m + 8 (m is Some Integer So m is 3q³+6q²+4q)

$\bold{\boxed{a3 = 9m + 8}}$

Hence It Is Proved That $\color{red}{the cube of Any Positive Integer is in The Form of 9m,9m+1,9m+8}$

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