Question -
Answer -
Solution:
Let x be any positive integer and y = 3.
By Euclid’s division algorithm
Now
x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0and r < 3.
Then, putting the value of r
We get,
x = 3q or x = 3q + 1 or x = 3q + 2
Now, by taking the cube of all the three aboveexpressions.
Case (i): Whenr = 0
Then,
x2= (3q)3 = 27q3=9(3q3)= 9m; where m = 3q3
Case (ii): Whenr = 1
Then,
x3 = (3q+1)3 =(3q)3 +13+3×3q×1(3q+1) = 27q3+1+27q2+9q
Taking 9 as common above factor
We get,
x3 = 9(3q3+3q2+q)+1
Putting = m
We get,
Putting (3q3+3q2+q) = m,we get ,
x3 = 9m+1
Case (iii): When r = 2
Then,
x3 = (3q+2)3
=(3q)3+23+3×3q×2(3q+2)
=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 = 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.