Using cubic root factor the expression $$a^2+3b^3$$ We have $$a^2+3b^3=\left(\sqrt[3]{a^2}\right)^3+\left(\sqrt[3]{3b^3}\right)^3 \tag{1} $$$$ =\left(\sqrt[3]{a^2}+\sqrt[3]{3b^3}\right)\left(\sqrt[3]{a^4}-\sqrt[3]{3a^2b^3}+\sqrt[3]{9b^6}\right)\tag{2} $$$$ =\left(\sqrt[3]{a^2}+b\sqrt[3]{3}\right)\left(a\sqrt[3]{a}-b\sqrt[3]{3a^2}+b^2\sqrt[3]{9}\right)\tag{3}$$
Am I right?
Equation $(2)$ does not expand or simplify to equation $(1).\quad$ It appears that the last term of equation $(2)$ is not right.
One way of factoring uses substitution in steps to make things easier to see.
\begin{align*} \space &a^2+3b^3=x^3+y^3\\ =\space &(x + y) (x^2 - x y + y^2)\\ =\space & \bigg( \sqrt[3]{a^2} + \sqrt[3]{3b^3}\bigg) \bigg( (\sqrt[3]{a^2}) ^2 - (\sqrt[3]{a^2}) (\sqrt[3]{3b^3}) + (\sqrt[3]{3b^3})^2\bigg)\\ =\space & \bigg( \sqrt[3]{a^2} + \sqrt[3]{3b^3}\bigg) \bigg( \big(\sqrt[3]{a^2}\big) ^2 - \sqrt[3]{3a^2b^3} + \big(\sqrt[3]{3b^3}\big)^2\bigg) \\ =\space &a^2+3b^3 \quad\text{for }\quad b\le 0 \space \text{ or }\space a^2+3b^3 > 0 \end{align*}
WoframAlpha supports this factoring here .
Also, here is an example showing the validity of these factors. Let
\begin{align*} f(a,b) &=\bigg( \sqrt[3]{a^2} + \sqrt[3]{3b^3}\bigg) &f(1,1) & = 1 + 3^{1/3} \\ g(a,b)&= \bigg( \sqrt[3]{a^2} ^2 - \sqrt[3]{3a^2b^3} + \sqrt[3]{3b^3}^2\bigg) &g(1,1) & = 1 - 3^{1/3} + 3^{2/3} \\ h(a,b)&=f\big(a,b\big) \cdot g\big(a,b\big) &h(1,1)& = \big(1+ 3^{1/3}\big)\\ &&&\space\cdot\space\big(1 - 3^{1/3} + 3^{2/3} \big) \\&&&=4\\ i(a,b)&=a^2+3b^3 &i(1,1) &=4 \\ \end{align*}
Another method assumes that the sum is zero, that $b$ is positive,
and the factors are complex.
\begin{align*} &a^2+3b^3=0\implies a=\pm\sqrt{-3b^3}\\ \space & a^2 - \big(\sqrt{3b^3}i\big)^2 \\ =\space & \bigg( a - \big(\sqrt{3b^3}i\big)\bigg) \bigg( a + \big(\sqrt{3b^3}i\big)\bigg) \end{align*}