Question

# If n be any positive integer, then by using Euclid's division algorithm, show that(i) n3 can be expressed in the form 9m,9m+1 or 9m+8.(ii) n3+1 can be expressed in the form 9m,9m+1 or 9m+2.Here, m is some integer.

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Solution

## Let n be any positive integer and b=3According to Euclid's division algorithm there exist integers q and r such that n=3×q+r where 0≤r<3 So value of r can be 0,1 and 2We can write n=3×q, n=3×q+1 and n=3×q+2Now take the each case one by one Case 1: n=3×qTake cube on both sides, we have n3=27×q3 = 9(3×q3)=9m where m=3q3∴n3+1=9m+1Similarly for 2nd case n=3×q+1Taking cube on both sides, we haven3=27q3+27q2+9q+1 or n3=9(3q3+3q2+q)+1Which is equal to 9m+1 where m=(3q3+3q2+q)Thus, n3+1=9m+2For final case n=3×q+2 Taking cube on both sides, we haven3=27q3+54q2+36q+8i.e. n3=9(3q3+6q2+4q)+8which is equal to 9m+8 where m=(3q3+6q2+4q)Hence, n3+1=9m+8+1=9m+9=9(m+1)=9n where n=m+1 and n is an integer

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