Verified answer

[tex]\displaystyle\bf\\\frac{8}{125} a^{12} \Bigg(\frac{5}{a} +\frac{1}{2a} \Bigg)^{3} =\\\\\\\frac{8}{125} a^{12} \Bigg(\frac{125}{a^{3} } +3\cdot\frac{25}{a^{2} } \cdot \frac{1}{2a} +3\cdot\frac{5}{a} \cdot \frac{1}{4a^{2} } +\frac{1}{8a^{3} } \Bigg)=\\\\\\=\frac{8}{125} a^{12} \Bigg(\frac{125}{a^{3} } +\frac{75}{2a^{3} } +\frac{15}{4a^{3} } +\frac{1}{8a^{3} } \Bigg)=\\\\\\=8a^{9} +2,4a^{9}+0,24a^{9} +0,008a^{9}[/tex]