Question

show that the cube of any possitive integer is in the form 4m or 4m+1 or 4m+3 where m is a whole number
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Answer 1

GIVEN:

• Then forms of numbers are

1. 4m
2. 4m + 1
3. 4m + 3

• m is a whole number.

TO PROVE:

• The cube of any possitive integer is in the form 4m or 4m + 1 or 4m + 3.

EXPLANATION:

By Euclid's division algorithm,

$\boxed{\bold{\pink{\large{a = bq + r}}}}$

•°• Here a be any positive integer, b = 4.

•°• Also q ≥ 0, Since a is positive.

•°• r ≤ 0 < 4, So r = 0, 1, 2, 3. [ As b = 4, r cannot exceed 4 ]

Possible values of a:

• a = 4q + 0

• a = 4q + 1

• a = 4q + 2

• a = 4q + 3

•°• Let us take a = 4q

Cubing on both sides.

•°• a³ = 64q³

•°• a³ = 4(16q³)

•°• It is in the form of 4m

•°• Here m = 16q³

•°• Let us take a = 4q + 1

Cubing on both sides.

$\boxed{\bold{\orange{\large{(A + B)^3 = A^3 + B^3 + 3A^2B + 3AB^2}}}}$

•°• a³ = 64q³ + 1 + 3(16q²)(1) + 3(4q)(1)²

•°• Take 4 as common

•°• a³ = 4(16q³ + 12q² + 3q) + 1

•°• It is in the form of 4m + 1

•°• Here m = 16(16q³ + 12q² + 3q)

•°• Let us take a = 4q + 2

Cubing on both sides.

$\boxed{\bold{\blue{\large{(A + B)^3 = A^3 + B^3 + 3A^2B + 3AB^2}}}}$

•°• a³ = 64q³ + 2³ + 3(16q²)(2) + 3(4q)(2)²

•°• a³ = 64q³ + 8 + 3(16q²)(2) + 3(4q)(2)²

•°• Take 4 as common

•°• a³ = 4(16q³ + 24q² + 12q + 2)

•°• It is in the form of 4m

•°• Here m = 16(16q³ + 24q² + 12q + 2)

•°• Let us take a = 4q + 3

Cubing on both sides.

$\boxed{\bold{\red{\large{(A + B)^3 = A^3 + B^3 + 3A^2B + 3AB^2}}}}$

•°• a³ = 64q³ + 3³ + 3(16q²)(3) + 3(4q)(3)²

•°• a³ = 64q³ + 27 + 3(16q²)(3) + 3(4q)(3)²

•°• a³ = 64q³ + 9(16q²) + 27(4q) + 24 + 3

•°• Take 4 as common

•°• a³ = 4(16q³ + 36q² + 27q + 6) + 3

•°• It is in the form of 4m + 3

•°• Here m = 16(16q³ + 36q² + 27q + 6)

Hence proved that the cube of any positive integer is in the form 4m or 4m + 1 or 4m + 3.