1. Use the cosine rule at angle $C$

Since $\angle ACB = \dfrac{\pi}{6}$ and the side opposite $C$ is $c$, we use

$c^2 = a^2 + b^2 - 2ab\cos\frac{\pi}{6}$

Now,

$\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$

So,

$c^2 = a^2 + b^2 - \sqrt{3}\,ab$

Given

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

Substitute these into the cosine rule.

2. Simplify using $a-b$

Notice:

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

Hence,

$a^2+b^2 = (a-b)^2 + 2ab = (x+2)^2 + 2ab$

Therefore,

$c^2 = (x+2)^2 + 2ab - \sqrt{3}\,ab$

So,

$c^2 = (x+2)^2 + (2-\sqrt{3})ab$

Now compute $ab$:

$ab = (x^2+x+1)(x^2-1)$

Factor:

$x^2+x+1 = \frac{x^3-1}{x-1} \quad \text{(not especially useful here)}$

Direct expansion:

$ab = x^4 + x^3 - x - 1$

Also,

$c^2 = (2x+1)^2 = 4x^2+4x+1$

and

$(x+2)^2 = x^2+4x+4$

Thus,

$4x^2+4x+1 = x^2+4x+4 + (2-\sqrt{3})(x^4+x^3-x-1)$

So,

$3x^2-3 = (2-\sqrt{3})(x^4+x^3-x-1)$

Factor the left side:

$3(x^2-1)=3(x-1)(x+1)$

Factor the quartic on the right:

$x^4+x^3-x-1 = x^3(x+1)-1(x+1)=(x+1)(x^3-1)$

$=(x+1)(x-1)(x^2+x+1)$

Hence,

$3(x-1)(x+1) = (2-\sqrt{3})(x+1)(x-1)(x^2+x+1)$

So,

$(x^2-1)\left[3-(2-\sqrt{3})(x^2+x+1)\right]=0$

3. Solve the equation

This gives two cases:

Case 1: $x^2-1=0$

$x=\pm 1$

Check whether these give valid triangle sides:

• If $x=1$: $a=3,\ b=0,\ c=3$ Not a triangle since $b=0$.

• If $x=-1$: $a=1,\ b=0,\ c=-1$ Invalid since side lengths cannot be non-positive.

So these are rejected.

Case 2:

$3=(2-\sqrt{3})(x^2+x+1)$

Therefore,

$x^2+x+1 = \frac{3}{2-\sqrt{3}}$

Rationalize:

$\frac{3}{2-\sqrt{3}} = \frac{3(2+\sqrt{3})}{4-3}=3(2+\sqrt{3})=6+3\sqrt{3}$

So,

$x^2+x+1 = 6+3\sqrt{3}$

$x^2+x-(5+3\sqrt{3})=0$

Now solve:

$x = \frac{-1\pm \sqrt{1+4(5+3\sqrt{3})}}{2}$

$x = \frac{-1\pm \sqrt{21+12\sqrt{3}}}{2}$

Notice,

$(3+2\sqrt{3})^2 = 9+12\sqrt{3}+12=21+12\sqrt{3}$

Hence,

$\sqrt{21+12\sqrt{3}} = 3+2\sqrt{3}$

Therefore,

$x = \frac{-1\pm (3+2\sqrt{3})}{2}$

So,

$x = 1+\sqrt{3} \quad \text{or} \quad x = -(2+\sqrt{3})$

4. Check validity of side lengths

• For $x=1+\sqrt{3}$: $c=2x+1=3+2\sqrt{3}>0, \quad b=x^2-1>0, \quad a>0$ Valid.

• For $x=-(2+\sqrt{3})$: $c=2x+1 = -3-2\sqrt{3}<0$ Invalid since side length cannot be negative.

So the only admissible value is

$x=1+\sqrt{3}$

5. Match with options

This corresponds to Option B.