A function f is defined by f(x) = x^3 + a x^2 + b x + c , where a, b and c are constants. The graph of y = f(x) passes through the points (-2, 1) and (2, -3). The point (2, 1) lies on the graph of y = f(x + 1). Find the values of a, b and c.

1. Form two equations using the points $(-2, 1)$ and $(2, -3)$ on the graph of $y = \mathrm{f}(x)$.

Since the graph of $y = \mathrm{f}(x)$ passes through $(-2, 1)$ and $(2, -3)$, we have $\mathrm{f}(-2) = 1$ and $\mathrm{f}(2) = -3$.

$(-2)^3 + a(-2)^2 + b(-2) + c = 1$
$-8 + 4a - 2b + c = 1$
$4a - 2b + c = 9 \quad \text{--- (1)}$

$(2)^3 + a(2)^2 + b(2) + c = -3$
$8 + 4a + 2b + c = -3$
$4a + 2b + c = -11 \quad \text{--- (2)}$

2. Form a third equation using the point $(2, 1)$ on the graph of $y = \mathrm{f}(x + 1)$.

The point $(2, 1)$ lies on $y = \mathrm{f}(x + 1)$, so substituting $x = 2$ and $y = 1$ yields:
$1 = \mathrm{f}(2 + 1) \implies \mathrm{f}(3) = 1$

Substituting $x = 3$ into $\mathrm{f}(x)$:
$3^3 + a(3)^2 + b(3) + c = 1$
$27 + 9a + 3b + c = 1$
$9a + 3b + c = -26 \quad \text{--- (3)}$

3. Solve the system of three linear equations to find $a$, $b$, and $c$.

Subtracting (1) from (2):
$(4a + 2b + c) - (4a - 2b + c) = -11 - 9$
$4b = -20 \implies b = -5$

Substituting $b = -5$ into (1) and (3):
$4a - 2(-5) + c = 9 \implies 4a + c = -1 \quad \text{--- (4)}$
$9a + 3(-5) + c = -26 \implies 9a + c = -11 \quad \text{--- (5)}$

Subtracting (4) from (5):
$(9a + c) - (4a + c) = -11 - (-1)$
$5a = -10 \implies a = -2$

Substituting $a = -2$ into (4):
$4(-2) + c = -1 \implies -8 + c = -1 \implies c = 7$

(Note: The system can also be solved directly using a graphing calculator.)