A cubic polynomial curve y = f(x) = ax^3 + bx^2 + cx + d has a...

Let $f(x) = ax^3 + bx^2 + cx + d$.

Since the tangent to the curve at $(0, 2)$ is $y = 3x + 2$, the curve passes through $(0, 2)$.
Thus, $f(0) = 2 \Rightarrow d = 2$.

The slope of the tangent at $x = 0$ is $3$.
$f'(x) = 3ax^2 + 2bx + c$
$f'(0) = 3 \Rightarrow c = 3$.

The curve has a point of inflection at $x = 1$, so $f''(1) = 0$.
$f''(x) = 6ax + 2b$
$f''(1) = 6a + 2b = 0 \Rightarrow 3a + b = 0 \Rightarrow b = -3a$.

The curve passes through $(1, 1)$, so $f(1) = 1$.
$a + b + c + d = 1$
$a + (-3a) + 3 + 2 = 1$
$-2a + 5 = 1 \Rightarrow -2a = -4 \Rightarrow a = 2$.

Then $b = -3(2) = -6$.

We have $a = 2$, $b = -6$, $c = 3$, $d = 2$.
Therefore, $a^2 + b^2 + c^2 + d^2 = 2^2 + (-6)^2 + 3^2 + 2^2 = 4 + 36 + 9 + 4 = 53$.

Answer: 53