let's recall that
1² = 1
1⁴ = 1
1¹⁰⁰⁰⁰⁰⁰⁰⁰⁰ = 1
[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}\\\\ x^{12}y^{18}+1\implies x^{4\cdot 3}y^{6\cdot 3}+1^3\implies (x^4)^3(y^6)^3+1^3\implies (x^4y^6)^3+1^3 \\[2em] [x^4y^6+1][(x^4y^6)^2-(x^4y^6)(1)+1^2]\implies (x^4y^6+1)(x^8y^{12}-x^4y^6+1)[/tex]
