For $$t > -1$$, let $$\alpha_t$$ and $$\beta_t$$ be the roots of the equation $$\left((t+2)^{1/7} - 1\right)x^2 + \left((t+2)^{1/6} - 1\right)x + \left((t+2)^{1/21} - 1\right) = 0$$. If $$\lim_{t \to -1^+} \alpha_t = a$$ and $$\lim_{t \to -1^+} \beta_t = b$$, then $$72(a + b)^2$$ is equal to _____.

Let $$A(t) = (t+2)^{1/7}-1,\; B(t) = (t+2)^{1/6}-1,\; C(t) = (t+2)^{1/21}-1$$.
For $$t \gt -1$$ the quadratic in $$x$$ is

$$A(t)\,x^{2}+B(t)\,x+C(t)=0$$

Put $$t=-1+\varepsilon$$ with $$\varepsilon \to 0^{+}$$, so $$t+2 = 1+\varepsilon$$.
Using the first-order binomial approximation $$\bigl(1+\varepsilon\bigr)^{k} = 1+k\varepsilon+O(\varepsilon^{2})$$ as $$\varepsilon \to 0$$, we get

$$A(t)= (1+\varepsilon)^{1/7}-1 = \tfrac{1}{7}\varepsilon+O(\varepsilon^{2}),$$
$$B(t)= (1+\varepsilon)^{1/6}-1 = \tfrac{1}{6}\varepsilon+O(\varepsilon^{2}),$$
$$C(t)= (1+\varepsilon)^{1/21}-1 = \tfrac{1}{21}\varepsilon+O(\varepsilon^{2}).$$

Hence the quadratic becomes

$$\bigl(\tfrac{1}{7}\varepsilon+O(\varepsilon^{2})\bigr)x^{2}+ \bigl(\tfrac{1}{6}\varepsilon+O(\varepsilon^{2})\bigr)x+ \bigl(\tfrac{1}{21}\varepsilon+O(\varepsilon^{2})\bigr)=0.$$

Because $$\varepsilon \neq 0$$, divide by $$\varepsilon$$ and then let $$\varepsilon \to 0^{+}$$:

$$\tfrac{1}{7}x^{2}+\tfrac{1}{6}x+\tfrac{1}{21}=0.$$

This limiting quadratic has roots $$a=\lim_{t\to-1^{+}}\alpha_{t}$$ and $$b=\lim_{t\to-1^{+}}\beta_{t}$$.

For a quadratic $$px^{2}+qx+r=0$$, the sum of the roots is $$-\dfrac{q}{p}$$.
Here $$p=\tfrac{1}{7},\; q=\tfrac{1}{6}$$, so

$$a+b = -\frac{\tfrac{1}{6}}{\tfrac{1}{7}} = -\frac{7}{6}.$$

Therefore

$$72\,(a+b)^{2} = 72 \left(-\frac{7}{6}\right)^{2} = 72 \cdot \frac{49}{36} = 2 \cdot 49 = 98.$$

Hence the required value is $$\boxed{98}$$.