\(64=\left(a^2+a+2\right)\left(b+1\right)^2\left(c^2+3c\right)\)
\(=\left(a^2+a+1+1\right)\left(b^2+b+b+1\right)\left(c^2+c+c+c\right)\)
\(\ge4.\sqrt[4]{a^3}.4.\sqrt[4]{b^4}.4.\sqrt[4]{c^5}\)
\(=64\sqrt[4]{a^3b^4c^5}\)
\(\Rightarrow\sqrt[4]{a^3b^4c^5}\le1\)
\(\Leftrightarrow a^3b^4c^5\le1\)