## 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: $(3^{3^{n-1}}+9^{3^{n-1}}-1)(9^{3^{n-1}}+81^{3^{n-1}}+1-27^{3^{n-1}}+3^{3^{n-1}}+9^{3^{n-1}})$

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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