${\cos }^{6}A+{\sin }^{6}A=$

${\left({\cos }^{2}A\right)}^3+{\left({\sin }^{2}A\right)}^3$

$\Rightarrow {(a+b)}^3-3ab(a+b)=a^3+b^3$

$={({\cos }^{2}A+{\sin }^{2}A)}^3-3{\cos }^{2}A\cdot {\sin }^{2}A({\cos }^{2}A+{\sin }^{2}A)$

$=1-3{\sin }^{2}A\cdot {\cos }^{2}A$

$=1-\frac{3}{4}(2^2{\sin }^{2}A\cdot {\cos }^{2}A)$

$=1-\frac{3}{4}{\left(\sin 2A\right)}^2$

$=1-\frac{3}{4}{\sin }^{2}2A$

$=1-\frac{3}{4}(1-{\cos }^{2}2A)$

$=\frac{4-3(1-{\cos }^{2}2A)}{4}$

$=\frac{1}{4}[4-3+3{\cos }^{2}2A]$

$=\frac{1}{4}[1+3{\cos }^{2}2A]$