Question

use euclid divissin lemma ,show that cube of any positive interger in the form of 7m 7m+1,7m+6​

Answer 1

let a^3 be a cube of any positive integer.

then a is of the form 7q,7q+1,7q+2,7q+3,7q+4,7q+5 or 7q+6.

if a=7q,

a^3=7q^3=7m wherem=49(q)^3

if a=7q+1,

a^3=(7q+1)^3

=7q^3+

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Answer 2

Answer:

Step-by-step explanation:

Given use euclid divissin lemma ,show that cube of any positive interger in the form of 7m 7m+1,7m+6​

According to euclid division lemma we have

a = bq + r

let us consider b as 7 and let r = 0, 1, 2, 3,......

When r = 0,

a = 7q + 0

a = 7q

Cubing both sides we get

a^3 = (7q)^3

a^3 = 343q^3

When r = 1

a = 7q + 1

Cubing both sides we get

a^3 = (7q + 1)^3

We know (a + b)^3 = a^3 + b^3 + 3ab(a + b)

                (7q + 1)^3 = (7q)^3 + 1^3 + 21q(7q + 1)

                                 = 343q^3 + 147q^2 + 21q + 1

                 So a^3 =  7( 49q^3 + 21q^2 + 3q) + 1

                       a^3 = 7m + 1

 When r = 6

 a = 7q + 6

Cubing both sides we get

a^3 = (7q + 6)^3

 We know (a + b)^3 = a^3 + b^3 +3ab(a + b)

                (7q + 6)^3 = (7q)^3 + 6^3 + 42q(7q + 6)

                                 = 7(49q^3 + 127q^2 + 108q) + 216

                          a^3 = 7m + 216

 So 7m + 6 does not form a cube.