Question

show that any positive number's cube is of form 9m, 9m+1, 9m+8​

Answer 1

Answer:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m =

Case 2: When a = 3q + 1,

a 3 = (3q +1) 3

a 3 = 27q 3 + 27q 2 + 9q + 1

a 3 = 9(3q 3 + 3q 2 + q) + 1

a 3 = 9m + 1

Where m is an integer such that m = (3q 3 + 3q 2 + q)

Case 3: When a = 3q + 2,

a 3 = (3q +2) 3

a 3 = 27q 3 + 54q 2 + 36q + 8

a 3 = 9(3q 3 + 6q 2 + 4q) + 8

a 3 = 9m + 8

Where m is an integer such that m = (3q 3 + 6q 2 + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Answer 2

Solution

Let a be any positive integer

b = 3

According to Euclid's division lemma,

$\mathsf{a = 3q + r}$

$\sf{where \: q \geqslant 0 \: and \: 0 \leqslant r \: < 3}$

$\sf{Therefore, r = 0 \: or \: 1 \: or \: 2}$

So, every number can be expressed in these 3 forms..

1) a= 3q

a³ = 27q³ = 9m

a³=9m

"Taking m=3q³"

2) a= 3q+1

a³ = 27q³+27q²+9q+1

a³= 9m+1

"Taking m= 3q³+3q²+q"

3) a=(3q+2)³

=27q³+ 54q²+36q+8

a³=9m+8

"Taking m=3q³+6q²+4q"