Use euclid 's division lemma to show that the cube of any positive integer is of the form 9m, 9m+1,9m+8.
Answers
Hey mate !
Step-by-step explanation:
• Let the positive integer be a which when divided by 3 gives q as quotient and r as remainder.
by Euclid's division lemma
a=bq+r
a=3q+r
where r=0,1,2
then,
a=3q
or
a=3q+1
or
a=3q+2
now,
• CASE i
a=3q
a³=(3q)³
a³=27q³
a³=9m( where m=3q³)
• CASEii........ using identity
(a+b)³=a³+3a²b+3ab²+b³
a=3q+1
a³=(3q+1)³
a³=27q³+9q²+9q+1
a³=9m(3q³+q²+q)+1
a³=9m+1
• CASE iii
a=3q+2
a³=(3q+2)³
a³=27q³+27q²+36q+8
a³=9m(where m=3q³+3q²+4q)+8
a³=9m+8
Hence , the cube of any +ve no. is of the form 9m , 9m +1 or 9m+8
is shown !!
Step-by-step explanation:
Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3
∴ r = 0,1,2 .
Therefore, every number can be represented as these three forms. There are three cases.
Case 1: When a = 3q,
Where m is an integer such that m =
Case 2: When a = 3q + 1,
a = (3q +1) ³
a = 27q ³+ 27q ² + 9q + 1
a = 9(3q ³ + 3q ² + q) + 1
a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .
Case 3: When a = 3q + 2,
a = (3q +2) ³
a = 27q³ + 54q² + 36q + 8
a = 9(3q³ + 6q² + 4q) + 8
a = 9m + 8
Where m is an integer such that m = (3q³ + 6q² + 4q)
Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.
Hence, it is proved .