1.

use division algorithm to show that the cube of any positive integer is of the form 9m or 9m+1 or 9m+8

Answer»

We know,
For every POSITIVE integer a and b there always exists another two integers q and r such that a=bq+r and 0<=r

Taking b=3
a=3q+0 or
a=3q+1 or
a=3q+2
Now,
a^3=(3q)^3
=27q^3
=9×3q^3
=9m[Taking 3q^3=m]

a^3=(3q+1)^3
=(3q)^3 + 3×(3q)^2×1 + 3×3q×1^2+1^3
=27q^3 + 27q^2 +9q+1
=9(3q^3+3q^2+q)+1
=9m+1[Taking 3q^3+3q^2+q=m]

a^3=(3q+2)^3
=(3q)^3 + 3×(3q)^2×2 + 3×3q×2^2+2^3
=27q^3 + 54q^2 +36q+8
=9(3q^3+6q^2+4q)+8
=9m+8[Taking 3q^3+6q^2+4q=m]



proved....


hope it helps u....


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