Prove: An odd number cubed is an odd number.
n +
(2n + 1)3 = [ 2 ]n3 + [ ]n2 + [
[](4n3 + 6n2 + 3n) +
= an odd

Let n be any odd integer. Then n = 2k + 1 for some integer k.

So

n ³ = (2k + 1)³

… = (2k)³ + 3 (2k)² + 3 (2k) + 1

… = 8k ³ + 12k ² + 6k + 1

… = 2 (4k ³ + 6k ² + 3k) + 1

is also odd, since 4k ³ + 6k ² + 3k is just another integer.

the answer is 15% decrease

example 2 – simplify: step 1: completely factor both the numerators and denominators of all fractions. step 2: change the division sign to a multiplication sign and flip (or reciprocate) the fraction after the division sign; essential you need to multiply by the reciprocal. step 3 : cancel or reduce the fractions.