Reciprocated terms

We have that $3x+\dfrac{1}{2x} =3 \implies 2x= \dfrac{6x-1}{3x}$ Let $p= 2x$ and $q =\frac{1} {3x}$ and we need to evaluate $64x^6+\dfrac{1}{729x^6} =p^6 +q^6 = (p^2)^3 +(q^2)^3 \\ E = \left(p^2 +q^2\right)\left(p^4 +q^4-p^2.q^2\right) =\left( \left(p+q\right)^2 -2pq\right)\left(\left(p^2+q^2\right)^2 -2p^2q^2-p^2q^2\right) \\ E = \left( \left(2x+\dfrac{1}{3x}\right)^2 -\dfrac{2.2x}{3x}\right)\left(\left(\left(p+q\right)^2- 2pq\right)^2-3p^2q^2\right)\\ E = \left(\left(2x+\dfrac{1}{3x}\right)^2 -\dfrac{2.2x}{3x}\right)\left(\left(\left(2x+\dfrac{1}{3x}\right)^2- \dfrac{2.2x}{3x} \right)^2-\dfrac{3.4x^2}{9x^2}\right)\cdots (1)$ Now plugging the value of $2x=\frac{6x-1}{3x}$ in equation 1 we get $E = \left(\left(\dfrac{6x-1}{3x} +\dfrac{1}{3x}\right)^2 -\dfrac{2.2}{3}\right)\left(\left(\left(\dfrac{6x-1}{3x} +\dfrac{1}{3x}\right)^2- \dfrac{2.2}{3} \right)^2-\dfrac{12}{9} \right)\\ E = \left(4-\dfrac{4}{3}\right)\left(\left(4-\dfrac{4}{3}\right)^2 -\dfrac{12}{9} \right) = \dfrac{8}{3} \left( \dfrac{8^2}{9} -\dfrac{12}{9}\right)=\dfrac{416}{27}$ Hence , value of $a+b=443$ .