Question -
Answer -
Solution:
Let x be any positive integer and y = 3.
By EuclidтАЩs division algorithm
Now
x = 3q+r, where qтЙе0 and r = 0, 1, 2, as r тЙе 0and r < 3.
Then, putting the value of r
We get,
x = 3q┬а or┬а x = 3q + 1┬а┬а┬а┬аor┬а┬а┬а┬а┬а x = 3q + 2
Now, by taking the cube of all the three aboveexpressions.
Case (i):┬аWhenr = 0
┬аThen,
x2= (3q)3┬а= 27q3=9(3q3)= 9m; where m = 3q3
Case (ii):┬аWhenr = 1
┬аThen,
x3┬а= (3q+1)3┬а=(3q)3┬а+13+3├Ч3q├Ч1(3q+1) = 27q3+1+27q2+9q
Taking 9 as common above factor
We get,
x3┬а= 9(3q3+3q2+q)+1
Putting = m
We get,
Putting (3q3+3q2+q) = m,we get ,
x3┬а= 9m+1
Case (iii): When r = 2
Then,
x3┬а= (3q+2)3
┬а┬а┬а =(3q)3+23+3├Ч3q├Ч2(3q+2)
┬а┬а┬а =27q3+54q2+36q+8
Taking 9 as common above factor
We get,
x3=9(3q3+6q2+4q)+8
Putting (3q3+6q2+4q) = m
We get ,
x3┬а= 9m+8
Therefore, from all the three cases explainedabove, it is proved that the cube of any positive integer is of the form 9m, 9m+ 1 or 9m + 8.