Question

Use Euclid division lemma to show that the cube of any positive integer is of the form of 9m,(9m+1),(9m+8)?

Answer 1

Hey

Here is your answer,

Let ‘a’ is a positive integer and b = 4

Therefore, according Euclid’s division lemma,

a=3q+r where 0<_r<_3

so a=0,1,2

when a=0

a=3q

a^3=(3q)^3

a^3=27q^3

a^3=9(3q^3)

a^3=9m where m=3q^3.

when r=1

a=3q+1

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

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

a^3=27q^3+1+27q^2+9q

a^3=9(3q^3+3q^2+q)+1

a^3=9m+1 where m=3q^3+3q^2+q

hence cube of any positive integer is of the form 9m or 9m+1 or 9m+8

Hope it helps you!

Answer 2

Let a is integer and b=3
r=0,1,2
Case1,When r=0
a=3
a cube=(3q )cube
27q cube
9(3q)cube
9q
Case2,r=1
a=3q+1
a cube=(3q+1)cube
27q cube+1 cube+27q square+9
9(3q cube+3q+2)+1
9m+1
Case2,r=2
a=3q+2
a cube=(3q+2)cube
27q cube+8+54q square+36q
9(3q cube +6q square+4q) +8
9m+8
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