Using Binomial Theorem, the given expression $\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$ can be expanded as

$\left[\left(3 x^{2}-2 a x\right)+3 a^{2}\right]^{3}$

$ = {\,^3}{C_0}{\left( {3{x^2} – 2ax} \right)^3} + {\,^3}{C_1}{\left( {3{x^2} – 2ax} \right)^2}\left( {3{a^2}} \right) + {\,^3}{C_2}\left( {3{x^2} – 2ax} \right){\left( {3{a^2}} \right)^2} + {\,^3}{C_3}{\left( {3{a^2}} \right)^3}$

$=\left(3 x^{2}-2 a x\right)^{3}+3\left(9 x^{4}-12 a x^{3}+4 a^{2} x^{2}\right)\left(3 a^{2}\right)+3\left(3 x^{2}-2 a x\right)\left(9 a^{4}\right)+27 a^{6}$

$=\left(3 x^{2}-2 a x\right)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+36 a^{4} x^{2}+81 a^{4} x^{2}-54 a^{5} x+27 a^{6}$

$=\left(3 x^{2}-2 a x\right)^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}$            ………..$(1)$

Again by using Binomial Theorem, we obtain

$\left(3 x^{2}-2 a x\right)^{3}$

$ = {\,^3}{C_0}{\left( {3{x^2}} \right)^3} – {\,^3}{C_1}{\left( {3{x^2}} \right)^2}(2ax) + {\,^3}{C_2}\left( {3{x^2}} \right){(2ax)^2} – {\,^3}{C_3}{(2ax)^3}$

$=27 x^{6}-3\left(9 x^{4}\right)(2 a x)+3\left(3 x^{2}\right)\left(4 a^{2} x^{2}\right)-8 a^{3} x^{3}$

$=27 x^{6}-54 a x^{5}+36 a^{2} x^{4}-8 a^{3} x^{3}$           …………$(2)$

From $(1)$ and $(2),$ we obtain

$\left(3 x^{2}-2 a x+3 a^{2}\right)^{3}$

$=27 x^{6}-54 a x^{5}+36 a^{2} x^{4}-8 a^{3} x^{3}+81 a^{2} x^{4}-108 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}$

$=27 x^{6}-54 a x^{5}+117 a^{2} x^{4}-116 a^{3} x^{3}+117 a^{4} x^{2}-54 a^{5} x+27 a^{6}$