1. Interpret the limit

We need to evaluate

$$L=\lim_{n\to\infty}\frac{1+\sqrt[3]{2}+\cdots+\sqrt[3]{n}}{n^{7/3}\left(\frac1{(an+1)^2}+\frac1{(an+2)^2}+\cdots+\frac1{(an+n)^2}\right)}$$

and use the condition $L=54$ to find possible values of $a$, with $|a|>1$.

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2. Asymptotic form of the numerator

Let

$$S_n=1+\sqrt[3]{2}+\cdots+\sqrt[3]{n}=\sum_{k=1}^n k^{1/3}.$$

Using the standard Riemann sum / power-sum asymptotic,

$$\sum_{k=1}^n k^{1/3}\sim \int_0^n x^{1/3}\,dx=\frac34 n^{4/3}.$$

So,

$$S_n\sim \frac34 n^{4/3}.$$

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3. Asymptotic form of the denominator sum

Let

$$T_n=\sum_{k=1}^n \frac1{(an+k)^2}.$$

Write

$$T_n=\sum_{k=1}^n \frac1{n^2\left(a+\frac{k}{n}\right)^2} =\frac1{n^2}\sum_{k=1}^n \frac1{\left(a+\frac{k}{n}\right)^2}.$$

Hence

$$n^{7/3}T_n=n^{7/3}\cdot \frac1{n^2}\sum_{k=1}^n \frac1{\left(a+\frac{k}{n}\right)^2} =n^{1/3}\left(\frac1n\sum_{k=1}^n \frac1{\left(a+\frac{k}{n}\right)^2}\right).$$

As $n\to\infty$,

$$\frac1n\sum_{k=1}^n \frac1{\left(a+\frac{k}{n}\right)^2} \to \int_0^1 \frac{dx}{(a+x)^2}.$$

Therefore,

$$n^{7/3}T_n\sim n^{1/3}\int_0^1 \frac{dx}{(a+x)^2}.$$

Now,

$$\int_0^1 \frac{dx}{(a+x)^2} =\left[-\frac1{a+x}\right]_0^1 =\frac1a-\frac1{a+1} =\frac1{a(a+1)}.$$

So,

$$n^{7/3}T_n\sim \frac{n^{1/3}}{a(a+1)}.$$

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4. Evaluate the given limit

Thus,

$$L=\lim_{n\to\infty}\frac{S_n}{n^{7/3}T_n} \sim \frac{\frac34 n^{4/3}}{\frac{n^{1/3}}{a(a+1)}} =\frac34 ა(a+1)n.$$

This seems to grow like $n$, so let us re-check carefully.

The denominator is

$$n^{7/3}\left(\sum_{k=1}^n \frac1{(an+k)^2}\right).$$

Since each term is of order $1/n^2$, and there are $n$ terms, the sum is of order $1/n$. Therefore the whole denominator is of order

$$n^{7/3}\cdot \frac1n=n^{4/3},$$

which matches the numerator. Good — so the previous simplification missed one factor of $n$.

Let us correct it:

$$T_n=\sum_{k=1}^n \frac1{(an+k)^2} =\sum_{k=1}^n \frac1{n^2\left(a+\frac{k}{n}\right)^2} =\frac1{n^2}\sum_{k=1}^n \frac1{\left(a+\frac{k}{n}\right)^2}.$$

Now,

$$\sum_{k=1}^n \frac1{\left(a+\frac{k}{n}\right)^2} \sim n\int_0^1 \frac{dx}{(a+x)^2}.$$

Hence

$$T_n\sim \frac1{n^2}\cdot n\int_0^1 \frac{dx}{(a+x)^2} =\frac1n\int_0^1 \frac{dx}{(a+x)^2} =\frac1n\cdot \frac1{a(a+1)}.$$

Therefore,

$$n^{7/3}T_n\sim n^{7/3}\cdot \frac1n\cdot \frac1{a(a+1)} =\frac{n^{4/3}}{a(a+1)}.$$

So,

$$L=\lim_{n\to\infty}\frac{\frac34 n^{4/3}}{\frac{n^{4/3}}{a(a+1)}} =\frac34 a(a+1).$$

Given $L=54$, we get

$$\frac34 a(a+1)=54.$$

So,

$$a(a+1)=72.$$

Thus,

$$a^2+a-72=0.$$

Factorizing,

$$(a+9)(a-8)=0.$$

Hence,

$$a=8 \quad \text{or} \quad a=-9.$$

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5. Check options

• A: $-6$ $(-6)(-5)=30\ne 72$ Not correct.

• B: $-7$ $(-7)(-6)=42\ne 72$ Not correct.

• C: $8$ $8\cdot 9=72$ Correct.

• D: $-9$ $(-9)(-8)=72$ Correct.

Also both satisfy $|a|>1$.

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6. Final answer

The correct options are:

$$\boxed{\text{C, D}}$$