Verified answer

[tex]\displaystyle\bf\\8a^{3} -\frac{1}{8}=\frac{1}{8} \cdot(64a^{3} -1)=\frac{1}{8} \cdot\Big[(4a)^{3} -1^{3} \Big]=\\\\\\=\frac{1}{8}\cdot(4a-1)\cdot\Big[(4a)^{2} +4a\cdot 1+1^{2} \Big]=\frac{1}{8}\cdot(4a-1)\cdot(16a^{2} +4a+1)[/tex]