Let $$\lambda^*$$ be the largest value of $$\lambda$$ for which the function $$f_\lambda(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48$$ is increasing for all $$x \in \mathbb{R}$$. Then $$f_{\lambda^*}(1) + f_{\lambda^*}(-1)$$ is equal to:

We are given $$f_\lambda(x) = 4\lambda x^3 - 36\lambda x^2 + 36x + 48$$ and need to find the largest $$\lambda$$ for which $$f_\lambda$$ is increasing for all $$x \in \mathbb{R}$$.

For $$f_\lambda$$ to be increasing for all $$x$$, we need $$f'_\lambda(x) \geq 0$$ for all $$x$$.

$$f'_\lambda(x) = 12\lambda x^2 - 72\lambda x + 36$$

If $$\lambda = 0$$: $$f'(x) = 36 > 0$$ for all $$x$$, so $$\lambda = 0$$ works. But we want the largest $$\lambda$$.

If $$\lambda < 0$$: The quadratic $$12\lambda x^2 - 72\lambda x + 36$$ opens downward (since $$12\lambda < 0$$), so it will eventually become negative. Hence $$\lambda < 0$$ does not work.

If $$\lambda > 0$$: The quadratic opens upward. For it to be non-negative for all $$x$$, we need the discriminant $$\leq 0$$.

Discriminant $$= (72\lambda)^2 - 4(12\lambda)(36) = 5184\lambda^2 - 1728\lambda$$

Setting discriminant $$\leq 0$$:

$$5184\lambda^2 - 1728\lambda \leq 0$$

$$\lambda(5184\lambda - 1728) \leq 0$$

$$\lambda(3\lambda - 1) \leq 0$$ (dividing by 1728)

Since $$\lambda > 0$$, we need $$3\lambda - 1 \leq 0$$, i.e., $$\lambda \leq \frac{1}{3}$$.

Therefore $$\lambda^* = \frac{1}{3}$$.

Now computing $$f_{1/3}(1) + f_{1/3}(-1)$$:

$$f_{1/3}(x) = \frac{4}{3}x^3 - 12x^2 + 36x + 48$$

$$f_{1/3}(1) = \frac{4}{3} - 12 + 36 + 48 = \frac{4}{3} + 72 = \frac{220}{3}$$

$$f_{1/3}(-1) = -\frac{4}{3} - 12 - 36 + 48 = -\frac{4}{3} + 0 = -\frac{4}{3}$$

$$f_{1/3}(1) + f_{1/3}(-1) = \frac{220}{3} - \frac{4}{3} = \frac{216}{3} = 72$$

The correct answer is Option D.