Question

Show that the cube of any positive integer is of the form 4m

Answer 1

Answer:

sorry the question is wrong since integral values could not satisfy the question

Answer 2

Answer:

Show that cube of any positive integer is in the form of 4m, 4m + 1, 4m + 3

Solution :-

We know that, a number is of the form of

a = bq + r, where a, b are positive integers and 0 ≤ r < b

Now, a number can be of the form of

a = 4q

a = 4q + 1

a = 4q + 2

or, a = 4q + 3

if, b is 4

So, for the first case :-

a = 4q

⇒ a³ = (4q)³

⇒ a³ = 64q³

⇒ a³ = 4(16q³)

⇒ a³ = 4m

where, m = 16q³

Second case :-

a = 4q + 1

⇒ a³ = (4q + 1)³

⇒ a³ = 64q³ + 48q² + 12q + 1

⇒ a³ = 4(16q³ + 12q² + 3q) + 1

⇒ a³ = 4m + 1

where, m = (16q³ + 12q² + 3q)

Third Case :-

a = 4q + 2

⇒ a³ = (4q + 2)³

⇒ a³ = 64q³ + 96q² + 48q + 8

⇒ a³ = 4(16q³ + 24q² + 12q + 2)

⇒ a³ = 4m

where, m = (16q³ + 24q² + 12q + 2)

Fourth Case :-

a = 4q + 3

⇒ a³ = (4q + 3)³

⇒ a³ = 64q³ + 576q² + 108q + 27

⇒ a³ = 64q³ + 576q² + 108q + 24 + 3

⇒ a³ = 4(16q³ + 144q² + 27q + 6) + 3

⇒ a³ = 4m + 3

where, m = (16q³ + 144q² + 27q + 6)

Hence Proved.

Step-by-step explanation:

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