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By euclids division lemma
A = bq +r, 0>=r >b.
Let b=3 then r= 0,1,2
Now when r=0 then
A=3q
A^3 = 27q^3
A^3 =9(3q^3)
A^3 = 9m where m= 3q^3.
When r= 1
A= 3q +1
A^3 =27q^3 +27q^2 +9q+1
A^3 = 9(3q^3 +3q^2 +q) +1
A^3 = 9m +1 where m= 3q^3 +3q^3 +q
When r=2
A= 3q+2
A^3 = 27q^3 +54q^2 +36q+8
A^3 =9(3q^3+6q^2 +4q) + 8
A^3 = 9m+8 where m= 3q^3+6q^2+4q
Hence proved cube of a no is in the form 9m, 9m+1 or 9m+8.
Shailesh183816:
Why we take value of b = 3
we Can take value of 4,5,6 another.......
please reply me
Because 3 is a factor of nine and by taking b=3 we get least possible no of r{only 0 1 2}
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Hey
Let a is an integer
On dividing a by 3 we get
Q is quotient and, r is remainder
THEN, r=0,1,2
a=3q+r
When r=0
a=3q (cube on both sides)
a*3=27q*3
a*3=9(3q*3)
a*3=9m
When r=1
a=3q+1 (cube)
a*3=(3q+1)*3
a*3=27q*3+1+27q*2+9q
a*3=9(3q*3+3q*2+q) +1
a*3=9m+1
When r=2
a*3=(3q+2)*3
a*3=27q*3+8+54q*2+36q
a*3=9(3q*3+6q*2+4q)+8
a*3=9m+8
It is proved that cube of any positive integer in the form of 9m,9m+1,9m+2
Let a is an integer
On dividing a by 3 we get
Q is quotient and, r is remainder
THEN, r=0,1,2
a=3q+r
When r=0
a=3q (cube on both sides)
a*3=27q*3
a*3=9(3q*3)
a*3=9m
When r=1
a=3q+1 (cube)
a*3=(3q+1)*3
a*3=27q*3+1+27q*2+9q
a*3=9(3q*3+3q*2+q) +1
a*3=9m+1
When r=2
a*3=(3q+2)*3
a*3=27q*3+8+54q*2+36q
a*3=9(3q*3+6q*2+4q)+8
a*3=9m+8
It is proved that cube of any positive integer in the form of 9m,9m+1,9m+2
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