Given that $\left|\begin{array}{lll}1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$, the value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ 2 & 3 & 4 \\ 8 & 27 & 64\end{array}\right|$ is

18

$\left|\begin{array}{lll}1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3\end{array}\right|=\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3\end{array}\right|$

In the problem a = 2, b = 3, c = 4

Hence (3) is the correct answer.