Devilish fun

We have $f(x)=(x-1)^4-667$ , so the real roots must satisfy $(x-1)^2=\sqrt{667}$ $x^2-2x+1-\sqrt{667}=0.$ By Vieta's, the product of the roots is $1-\sqrt{667}$ , so $a+b=\boxed{668}$ .

A nice tricky problem. Observe that $f'(x) = 4(x^3 - 3x^2 + 3x - 1) = 4(x-1)^3$

Integrating both sides & using $f(0)=-666$ , we get: $f(x) = (x-1)^4 - 667$

At $f(x)=0$ , $(x-1)^4 = 667 \Rightarrow (x-1)^2 = \pm \sqrt{667}$

We neglect $(x-1)^2 = - \sqrt{667}$ ,as it contributes to complex roots & take $(x-1)^2 = \sqrt{667}$ , whose roots are the desired reals. Expanding, $x^2 - 2x +(1-\sqrt{667}) = 0$ . So product of the required roots is $1-\sqrt{667}$ .

When we examine the equation, we see that it looks very similar to $(x-1)^4=x^4-4x^3+6x^2-4x+1$ .

Therefore, we add 667 to both sides to get $x^4-4x^3+6x^2-4x+1=667 \implies (x-1)^4=667 \implies x-1= \sqrt[4]{667} \implies x=1 \pm \sqrt[4]{667}$

Therefore, when we multiply $(1+\sqrt[4]{667}) (1-\sqrt[4]{667})$ , we get $1-\sqrt{667} \text{ Which implies that our final answer is } \boxed{668}$

First notice that the first 4 terms are the expansion of the first 4 terms in (x + 1)^4. We can rewrite the function as (x + 1)^4 - 667 = 0. Then (x + 1)^4 = 667. Take the square root of both sides which gives (x + 1)^2 = sqrt(667). The two non real solutions are found when (x + 1)^2 = -sqrt(667). Since the question asks for the product of the real roots we only need to look at the first equation, (x + 1)^2 = sqrt(667). This tells you that the two values of x are 1 - 4th root(667) and 1 + 4th root(667). The product of the two solutions is 1 - sqrt(667). Hence, the sum of a and b is 668.

Looking at the coefficients of $f(x)$ one can feel that the polynomial is symmetric, indeed the polynomial is symmetric about the $x = 1$ axis. Hence $f(1 - x) = f(1 + x)$ How we found the axis of symmetry? Well, note that $f'(x) = 4(x - 1)^3$ Hence $f(x)$ attains minimum if and only if $x = 1$ , checking that it is also axis of symmetry is trivial. And now we just proceed easily noting that $\begin{aligned} f(1 - x) &= x^4 - 667 \\ &= \left(x - \sqrt[4]{667}\right) \cdot \left(x + \sqrt[4]{667}\right) \cdot \left(x^2 + \sqrt{667}\right) \end{aligned}$ This has roots $\pm \sqrt[4]{667}$ and $\pm i\sqrt[4]{667}$ . Hence $f(x)$ has roots $1 \pm \sqrt[4]{667}$ , $1 \pm i\sqrt[4]{667}$ , so the product of the real roots is $\left(1 - \sqrt[4]{667}\right) \cdot \left(1 + \sqrt[4]{667}\right) = 1 - \sqrt{667}$ Hence $a + b = 1 + 667 = \boxed{668}$