Minimum and maximum value

This is very similar to one of my questions, and the key observation to make is the following equality:

$a^2 - \frac{(a+1)^2 (8a - 1)}{27} = \frac{(1 - 2a)^3}{27}$

Using the identity $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ and expanding $f(a)^3$ gives

$f(a)^3 = 2a + (1-2a) f(a)$

Therefore, $f(a)$ is a real root of the equation $x^3 + (2a - 1)x - 2a = 0$ , which clearly has $x = 1$ as a real root. Dividing out by $x-1$ gives $(x-1)(x^2 + x + 2a)$ , and the discriminant of the quadratic is $1 - 8a$ , which is less than 0 when $a > 1/8$ , so in these cases we must have $f(a) = 1$ . For $a = 1/8$ , we have that $f(a) = 1$ by direct substitution, therefore $f(x) = 1$ is the constant function. The answer is therefore $100 - 31 = 69$ .

Let $b = \dfrac{8a-1}{3} \implies a = \dfrac{3b+1}{8}$

$\begin{aligned} a + \frac{a+1}{3}\sqrt{\frac{8a-1}{3}} & = \frac{3b+1}{8} + \frac{1}{3}\left(\frac{3b+1}{8} + 1 \right) \sqrt{b} \\ & = \frac{1}{8} \left(3b+1 + (b+3)\sqrt{b} \right) \\ & = \frac{1}{8} \left( b\sqrt{b}+ 3b + 3\sqrt{b} + 1 \right) \\ & = \frac{1}{8} \left( (\sqrt{b})^3+ 3(\sqrt{b})^2 + 3\sqrt{b} + 1 \right) \\ & = \frac{1}{8} \left(1 + 3\sqrt{b} \right)^3 \end{aligned}$

Similarly,

$\begin{aligned} a - \frac{a+1}{3}\sqrt{\frac{8a-1}{3}} & = \frac{1}{8} \left(1 - 3\sqrt{b} \right)^3 \end{aligned}$

Therefore, we have:

$\begin{aligned} f(a) & = \sqrt[3]{a + \frac{a+1}{3}\sqrt{\frac{8a-1}{3}}} + \sqrt[3]{a - \frac{a+1}{3}\sqrt{\frac{8a-1}{3}}} \\ & = \sqrt[3]{\frac{1}{8} \left(1 + 3\sqrt{b} \right)^3} + \sqrt[3]{\frac{1}{8} \left(1 - 3\sqrt{b} \right)^3} \\ & = \frac{1}{2}\left(1 + 3\sqrt{b} \right) + \frac{1}{2}\left(1 - 3\sqrt{b} \right) \\ & = 1 \quad \color{#3D99F6}{\text{(a constant)}} \\ \implies x & = y = 1 \end{aligned}$

$\implies 100x-31y = \boxed{69}$