Use Euclid’s division lemma to show that cube of any positive integer is of the form
9m, 9m + 1, 9m + 8.
Answers
Answer:
Let a and b be two positive integers, and a>b
a=(b×q)+r where q and r are positive integers and
0≤r<b
Let b=3 (If 9 is multiplied by 3 a perfect cube number is obtained)
a=3q+r where 0≤r<3
(i) if r=0,a=3q (ii) if r=1,a=3q+1 (iii) if r=2,a=3q+2
Consider, cubes of these
Case (i) a=3q
a3=(3q)3=27q3=9(3q3)=9m where m=3q3 and 'm' is an integer.
Case (ii) a=3q+1
a3=(3q+1)3 [(a+b)3=a3+b3+3a2b+3ab2]
=27q3+1+27q2+9q=27q3+27q2+9q+1
=9(3q
Let us consider a and b where a be any positive number and b is equal to 3
According to Euclid’s Division Lemma
a = bq + r
where r is greater than or equal to zero and less than b
(0 ≤ r < b)
a = 3q + r
so r is an integer greater than or equal to 0 and less than 3.
Hence r can be either 0, 1 or 2.
Case 1: When r = 0, the equation becomes
a = 3q
Cubing both the sides
Case 2: When r = 1, the equation becomes
a = 3q + 1