Some non trivial factorization examples

I hope to give you a flavour of some non-trivial factorization examples using the following identity:

a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab -bc-ca) Example 1. Let n be a positive integer. Factorize

3^{3^{n}}(3^{3^{n}}+1) +3^{3^{n}+1}-1

Solution. Observe that

3^{3^{n}}(3^{3^{n}}+1) +3^{3^{n}+1}-1 = a^{3}+b^{3}+c^{3}-3abc where a=3^{3^{n-1}}, b=9^{3^{n-1}}, and c=-1.

Thus, using the above factorization identity, we get the following factorization:


Example 2. Let a, b, c be distinct positive integers and let k be a positive integer such that ab+bc+ca \geq 3k^{2}-1.

Prove that (1/3)(a^{3}+b^{3}+c^{3})-abc \geq 3k.

Solution. The desired inequality is equivalent to

a^{3}+b^{3}+c^{3}-3abc \geq 9k.

Suppose without loss of generality, that a>b>c.

Then, since a, b, and c are distinct positive integers, we a-b \geq 1, and

(b-c) \geq 1 and a-c \geq 2.

It follows that a^{2}+b^{2}+c^{2}-ab-bc-ca = (1/2)((a-b)^{2}+(b-c)^{2}+(c-a)^{2}) \geq (1/2)(1+1+4)=3

We obtain

a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca) \geq 3(a+b+c) so it suffices to prove that 3(a+b+c) \geq 9k or (a+b+c) \geq 3k

But, (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2bc+2ca = a^{2}+b^{2}+c^{2} - ab -bc- ca +3(ab+bc+ca) \geq 3+3(3k^{2}-1) = 9k^{2},

and the conclusion follows.

More later… Nalin

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