Use euclids division lemma to show that the cube of any positive integer is in the form of 9m,9m+1, or 9m+8
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Let a be the no.
On dividing it by 3 , we get q as quotient and r as remainder.
By Euclid's Lemma, a=3q+r (0<_r<3)
a^3=(3q+r)^3
a^3=(27q^3+ r^3+9qr) (3q+r)
a^3=27q^3+r^3+27q^2r+9qr^2
a^3=9(3q^3+3q^2+qr^2)+r^3
where r=0
a=27q^3
=9(3Q^3)
=9m, where m= 3q^3
where r=1
a=9(3q^3+3q^2+q)+1
=9m+1
where r=2
9=(3q^3+6q^2+24q) +8
=9m+8
On dividing it by 3 , we get q as quotient and r as remainder.
By Euclid's Lemma, a=3q+r (0<_r<3)
a^3=(3q+r)^3
a^3=(27q^3+ r^3+9qr) (3q+r)
a^3=27q^3+r^3+27q^2r+9qr^2
a^3=9(3q^3+3q^2+qr^2)+r^3
where r=0
a=27q^3
=9(3Q^3)
=9m, where m= 3q^3
where r=1
a=9(3q^3+3q^2+q)+1
=9m+1
where r=2
9=(3q^3+6q^2+24q) +8
=9m+8
Answered by
1
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 .
THANKS.
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