Question

Show that the cube of any positive integer is of the form 9m,9m+1 or 9m+8

Answer 1

Answer:---

Let 'x' be any positive integer .

then , it is in the form of 3q , 3q+1,3q+2

$(3q) {3} = 27q {3} = 9 \times 3q {3}$

=9m (where m =3q)

$(3q + 1) {3} = 27q {3} +27q ^{2} + 9q + 1$

$= 9(3q {3} + 3q^{2} + q) + 1$

=9m+1. (where m=3q {3}^ + 3q ^{2} + q)

$(3q + 2) = 27q {3} + 54q^{2} + 36 + 8$

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

$( 3 q+ 2) {3} = 9m + 8$

(where m= 3q {3} + 6q ^{2} + 4q)

Answer 2

Answer:

step-by-step explanation:

Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0,1,2 .

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 = (3q +1) ³

a = 27q ³+ 27q ² + 9q + 1

a = 9(3q ³ + 3q ² + q) + 1

a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .

Case 3: When a = 3q + 2,

a = (3q +2) ³

a = 27q³ + 54q² + 36q + 8

a = 9(3q³ + 6q² + 4q) + 8

a = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

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

Hence, it is proved .