SAT 1000 problems - P6

$f(x) - f(a) = (x^3 + 3x^2 + 1) - (a^3 + 3a^2 + 1) = x^3 + 3x^2 - a^3 - 3a^2$ and $(x - b)(x - a)^2 = x^3 + (-2a - b)x^2 + (a^2 + 2ab)x - a^2b$

Since the two expressions are equal, $-2a - b = 3$ from the $x^2$ terms which means $2a + b = -3$ , and $a^2 + 2ab = 0$ from the $x$ terms which means $a + 2b = 0$ .

Adding these two equations gives $3a + 3b = -3$ which means $a + b = \boxed{-1}$ .

$\begin{aligned} f(x) - f(a) & = (x-b)(x-a)^2 \\ x^3 - a^3 + 3(x^2-a^2) & = (x-b)(x-a)^2 \\ (x-a)(x^2+ax+a^2) + 3(x-a)(x+a) & = (x-b)(x-a)^2 \\ x^2+ax+a^2 + 3(x+a) & = (x-b)(x-a) \\ x^2 + (a+3)x+a(a+3) & = x^2 -(a+b)x + ab \end{aligned}$

Equating the coefficients on both sides:

$\begin{cases} ab = a(a+3) & \implies b = a+3 \\ -(a+b) = a+3 & \implies -(a+a+3) = a+3 \implies a = -2 \implies b = 1 \end{cases}$

Therefore, $a+b = -2+1 = \boxed{-1}$ ,

{f(x) - f(a)} / (x-a) = (x-b) (x-a). Letting, x ->a yields f'(a) = 0. But a calculation shows f'(a) = 3a (a+2). Since a is non-zero, this means a = -2. We also know f(0) - f(a) = (0-b) (0-a)^2 = -b (4). Using the definition of f, f(0) - f(a) = f(0) - f(-2) = 1-5 = -4. Thus, -4*b = -4, or b = 1. Therefore, a+b = -2+1 = -1