Factorise

$64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}=(4 a)^{3}-(3 b)^{3}-3(4 a)^{2}(3 b)+3(4 a)(3 b)^{2}$

$=(4 a-3 b)^{3}$

Hence, factorisation of $64 a^{3}-27 b^{3}-144 a^{2} b+108 a b^{2}$ is $(4 a-3 b)^{3}$.