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