Question

show that the cube of any positive integer is of the form 4m, 4m+1 ,4m + 3 for some integer m.

Answer 1

Solution:

Let a be the positive integer and b = 4.

Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.

So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.

(4q)3 = 64q3 = 4(16q3)

= 4m, where m is some integer.

(4q + 1)3 = 64q3 + 48q2 + 12q + 1

= 4(16q3 + 12q2 + 3) + 1

= 4m + 1, where m is some integer.

(4q + 2)3 = 64q3 + 96q2 + 48q + 8

= 4(16q3 + 24q2 + 12q + 2)

= 4m, where m is some integer.

(4q + 3)3 = 64q3 + 144q2 + 108q + 27

= 4(16q3 + 36q2 + 27q + 6) + 3

= 4m + 3, where m is some integer.

Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

Answer 2

$\bold{Answer:-}$


$\bold{Step\: by \:step \:explanation}$



Using

$(a + b) {}^{3} = a {}^{3} + b {}^{3} + 3ab {}^{2} + 3a {}^{2} b$




⏩⏩Refer to the attachment above for your solution ⬆ ⬆






Hope it helps!