1. Given limit

We need to evaluate

$$L=\lim_{x\to\infty} \frac{(\sqrt{3x+1}+\sqrt{3x-1})^6+(\sqrt{3x+1}-\sqrt{3x-1})^6} {\left(x+\sqrt{x^2-1}\right)^6+\left(x-\sqrt{x^2-1}\right)^6}\,x^3$$

---

2. Use the identity

For any $a,b$,

$$(a+b)^6+(a-b)^6=2\left(a^6+15a^4b^2+15a^2b^4+b^6\right)$$

We apply this to the numerator with

$$a=\sqrt{3x+1},\qquad b=\sqrt{3x-1}$$

Then

$$a^2=3x+1,\qquad b^2=3x-1$$

So

$$(a+b)^6+(a-b)^6=2\left((3x+1)^3+15(3x+1)^2(3x-1)+15(3x+1)(3x-1)^2+(3x-1)^3\right)$$

But an easier way is to first compute

$$a^2+b^2=(3x+1)+(3x-1)=6x$$ $$a^2b^2=(3x+1)(3x-1)=9x^2-1$$

Using

$$(a+b)^6+(a-b)^6=2\left((a^2+b^2)^3-3a^2b^2(a^2+b^2)+12a^2b^2(a^2+b^2)\right)$$

which simplifies to

$$2\left((a^2+b^2)^3+12a^2b^2(a^2+b^2)\right)$$

Hence numerator

$$N=2\left((6x)^3+12(9x^2-1)(6x)\right)$$ $$N=2\left(216x^3+72x(9x^2-1)\right)$$ $$N=2\left(216x^3+648x^3-72x\right) =2(864x^3-72x)$$ $$N=1728x^3-144x$$

---

3. Simplify the denominator

Let

$$p=x+\sqrt{x^2-1},\qquad q=x-\sqrt{x^2-1}$$

Then

$$pq=x^2-(x^2-1)=1$$ $$p+q=2x$$

We need

$$D=p^6+q^6$$

Now,

$$p^3+q^3=(p+q)^3-3pq(p+q)=(2x)^3-3(1)(2x)=8x^3-6x$$

Therefore,

$$p^6+q^6=(p^3+q^3)^2-2p^3q^3$$

Since $pq=1$, we get $p^3q^3=1$. Thus

$$D=(8x^3-6x)^2-2$$ $$D=64x^6-96x^4+36x^2-2$$

---

4. Substitute into the limit

So

$$L=\lim_{x\to\infty} \frac{1728x^3-144x}{64x^6-96x^4+36x^2-2}\,x^3$$ $$L=\lim_{x\to\infty} \frac{(1728x^3-144x)x^3}{64x^6-96x^4+36x^2-2}$$ $$L=\lim_{x\to\infty} \frac{1728x^6-144x^4}{64x^6-96x^4+36x^2-2}$$

Divide numerator and denominator by $x^6$:

$$L=\lim_{x\to\infty} \frac{1728-\frac{144}{x^2}}{64-\frac{96}{x^2}+\frac{36}{x^4}-\frac{2}{x^6}}$$

As $x\to\infty$,

$$L=\frac{1728}{64}=27$$

---

5. Check options

• A: $9$ ❌
• B: $\frac{27}{2}$ ❌
• C: does not exist ❌
• D: $27$ ✅

Therefore, the correct answer is

$\boxed{27}$ which corresponds to Option D.