Question

show that cube of any number is 4q

Answer 1

Answer:

Step-by-step explanation:

Let 'a' be any positive integer, then it is of the form 4q or 4q+1 or 4q+2 4q+3 for some integer q

Case1: (4q)3 = 64q3 = 4 (16q3) where m = 16q3

Case 2: (4q + 1)3 = (4q)3 + 1+ 3 (16q2) + 3 (4q) = 64q3+ 48q2+ 12q +1 = 4q (16q2+ 12q + 3) + 1 where m = q (16q2 + 12q+ 3) for 4m+ 1

Case 3: (4q+ 2)3 = (4q)3+ 8+ 3 (4q)2 (2) + 3 (4q) (4) = 64q3+ 96q2+ 48q + 8 = 4 (16q3+ 24q2+ 12 + 2) where m= (16q3+ 24q2+ 12 + 2) for 4m

Case 4: (4q+3)3= (4q)3 + 27 + 3 (4q)2 (3) + 3 (4q) (9)= 64q3+ 36q2+ 108q+ 24+3 = 4 (16q3+ 9q2+ 27q+ 6) + 3 where m =(16q3+ 9q2+ 27q+ 6) for 4m+3

Hence cube of any positive integer is of the form 4m or 4m+1 or 4m+3

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