2120.12 – Factor!

Factor the number $6^{12} + 8^8 + 27^8$ without using $1$ as a factor.

Solution

\begin{aligned} 6^{12} + 8^8 + 27^8 &=& (2 \cdot 3)^{12} + (2^3)^8 + (3^3)^8 \hspace{1cm} [\approx 2.84623096 \times 10^{11}] \\ &=& 2^{12} \cdot 3^{12} + 2^{24} + 3^{24} \\ &=&2^{24} + 2^{12} \cdot 3^{12} + 3^{24} = 2^{24} + 2 (2^{12} \cdot 3^{12}) + 3^{24} - 2^{12} \cdot 3^{12} \\ &=& (2^{12})^2 + 2 (2^{12} \cdot 3^{12}) + (3^{12})^2 - 2^{12} \cdot 3^{12} \\ &=& (2^{12} + 3^{12})^2 - (2^6 \cdot 3^6)^2 \hspace{1cm} [2^{12} \cdot 3^{12} = (2^6 3^6)^2] \\ &=& (2^{12} + 3^{12} + 2^6 \cdot 3^6) (2^{12} + 3^{12} - 2^6 \cdot 3^6) \\ &=& 582193 \cdot 488881. \hspace{2cm} [\approx 2.84623096 \times 10^{11}] \end{aligned}