Question

Use euclids division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8 I want clear and detailed answer steps and no spamming please

Answer 1

According to Euclid’s Division Lemma

Let take a as any positive integer and b = 9.

Then using Euclid’s algorithm we get  a = 9q + r

  here r is remainder and value of q is more than or equal to 0  and r = 0, 1, 2, 3, 4, 5 , 6 , 7 , 8,
because 0 ≤r < b  and the value of b is 9

Sp possible forms will  9q, 9q+1, 9q+2,9q+3,9q+4,9q+5,9q+6,9q+7 and 9q+8

to get the cube of these values use the formula
(a+b)³           = a³ + 3a²b+ 3ab² + b³
In this formula value of a is always 9q
So plug the value we get
(9q+b)³          = 729q³ + 243q²b + 27qb² + b³

Now divide by 9 we get quotient = 81q³ + 27q²b + 3qb²  and remainder is b³

So we have to consider the value of b³
b = 0                     we get 9m+0  = 9m
b = 1                     then  1³  = 1 so we get  9m +1
b = 2                     then 2³ = 8 so we get 9m + 8
b = 3                     then 3³ = 27  and it is divisible by 9  so we get 9m
b = 4                     then 4³ =64 divide by 9 we get 1 as remainder so we get 9m +1
b=5                       then 5³=125 divide by 9 we get 8 as remainder so we get 9m+8
b=6                       then 6³=216 divide by 9 no remainder there so we get 9m
b=7                       then 7³ = 343 divide by 9 we get 1 as remainder so we get 9m+1
b=8                       then 8³ = 512 divide by 9 we get 8 as remainder so we get 9m+8
So all values are in form of 9m , 9m+1 or 9m+8 

Answer 2

According to Euclid's division lemma any two positive integers, say a and b, can be expressed as
a=bq+r
where 0<=r<b

Let's consider a position integer "a" and the value of b=3 (for this proof).

Nowas per Euclid's lemma "a" can be written as
a=3q+r
where 0<=r < 3

now we have three different cases:
Case 1: When r=0
So, a=3q only and by taking cube both sides we get
a³=(3q)³ = 27q³ = 9x3q³ = 9m
so the positive integer "a" can be written in form 9m where m=3q³

Case 2 : When r=1
So, a=3q+1 and by taking cube both sides we get
a³=(3q+1)³
= 27q³+27q²+9q+1
= 9(3q³+3q²+q)+1
=9m+1
so the positive integer "a" can be written in form 9m+1 where m=3q³+3q²+q

Case 3: when r = 2
So, a=3q+2 and by taking cube both sides we get
a³=(3q+2)³ = 27q³+54q²+36q+8
=9(3q³+6q²+4q)+8
=9m+8

so the positive integer "a" can be written in form 9m+8 where m=3q³+6q²+4q

Hence prooved