Question

What is the factorization of 729^15+1000​

Answer 1

`\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \begin{cases} 729=27^2\\ \qquad (3^3)^2\\ 1000=10^3 \end{cases}\implies 729^(15)+1000\implies ((3^3)^2)^(15)+10^3 \\\\\\ ((3^2)^(15))^3+10^3\implies (3^(30))^3+10^3\implies (3^(30)+10)~~[(3^(30))^2-(3^(30))(10)+10^2] \\\\\\ (3^(30))^3+10^3\implies (3^(30)+10)~~~~[(3^(60))-(3^(30))(10)+10^2]`

now, we could expand them, but there's no need, since it's just factoring.