Respuesta :

[tex]\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][/tex]

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