$$\mathop {\lim }\limits_{n \to \infty } \left[ {{{1 + \root 3 \of 2 + \root 3 \of 3 + ...\root 3 \of n } \over {{n^{7/3}}\left( {{1 \over {{{(an + 1)}^2}}} + {1 \over {{{(an + 2)}^2}}} + ... + {1 \over {{{(an + n)}^2}}}} \right)}}} \right]$$, $$a \in R,\,|a|\, > 1$$

$$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {({r^{1/3}})} } \over {{n^{7/3}}\sum\limits_{r = 1}^n {{1 \over {{{(an + r)}^2}}}} }}$$

$$ = \mathop {\lim }\limits_{n \to \infty } {{\sum\limits_{r = 1}^n {{{\left( {{r \over n}} \right)}^{1/3}}{1 \over n}} } \over {\sum\limits_{r = 1}^n {{1 \over {{{(a + {r \over n})}^2}}}{1 \over n}} }}$$

$$ = {{\int\limits_0^1 {{x^{1/3}}dx} } \over {\int\limits_0^1 {{{dx} \over {{{(a + x)}^2}}}} }} = 54$$, (given)

$$ \Rightarrow {{{3 \over 4}[{x^{4/3}}]_0^1} \over {\left[ { - {1 \over {x + a}}} \right]_0^1}} = 54$$

$$ \Rightarrow {{{3 \over 4}} \over { - {1 \over {a + 1}} + {1 \over a}}} = 54$$

$$ \Rightarrow {3 \over {4 \times 54}} = {1 \over {a(a + 1)}} $$

$$\Rightarrow {a^2} + a = 72$$

$$ \Rightarrow {a^2} + 9a - 8a - 72 = 0$$

$$ \Rightarrow a(a + 9) - 8(a + 9) = 0$$

$$ \Rightarrow (a - 8)(a + 9) = 0$$

$$ \Rightarrow a = 8$$ or $$ - 9$$

Hence, options (c) and (d) are correct.