# Prove that $512^3 + 675^3 + 720^3$ is a composite number

We have to prove that the number

$$N=512^3 + 675^3 + 720^3$$

is composite.

I tried to use the identity $(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$ hoping to take out some common factors from the R.H.S. but it didn't work. I also used $a^3+b^3=(a+b)(a^2-ab+b^2)$ (in all possible combos) and tried to combine with $c^3$ but that too didn't work. I nearly spent about 5 hours struggling with the question but no result :(
Let $a=512, b=675, c=720$. Now the number looks like $a^3+b^3+c^3$ but we require a sort of $3abc$ term to resolve it into factors. First we factorize $a,b,c$ into prime factors. So, $a=2^9, b=3^3\times 5^2, c=2^4\times 3^2\times 5$. Now it can be seen that $2c^2=3ab$. Hence, $a^3+b^3+c^3=a^3+b^3-c^3+2c^2c$ and thus the problem can be solved. It can then be seen that the given number is composite.