Question

use euclids division lemma to show that the cube of any positive integer is of the form of 3p or 3p+1 (or) 3p+2​

Answer 1

$\huge \star \red{answer}$

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.

Case II : When x = 3q + 1

then, x3 = (3q + 1)3

= 27q3 + 27q2 + 9q + 1

= 9 q (3q2 + 3q + 1) + 1

= 9m + 1, where m = q (3q2 + 3q + 1)

Case III. When x = 3q + 2

then, x3 = (3q + 2)3

= 27 q3 + 54q2 + 36q + 8

= 9q (3q2 + 6q + 4) + 8

= 9 m + 8, where m = q (3q2 + 6q + 4)

Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8.

Answer 2

Step-by-step explanation:

\huge \star \red{answer}⋆answer

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x3 = (3q)3 = 27q3 = 9 (3q3) = 9m, where m = 3q3.

Case II : When x = 3q + 1

then, x3 = (3q + 1)3

= 27q3 + 27q2 + 9q + 1

= 9 q (3q2 + 3q + 1) + 1

= 9m + 1, where m = q (3q2 + 3q + 1)

Case III. When x = 3q + 2

then, x3 = (3q + 2)3

= 27 q3 + 54q2 + 36q + 8

= 9q (3q2 + 6q + 4) + 8

= 9 m + 8, where m = q (3q2 + 6q + 4)

Hence, x3 is either of the form 9 m or 9 m + 1 or, 9 m + 8.