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

1. Form equations using the points given on the graph of $y = \text{f}(x)$.

Since $(-2, 1)$ lies on $y = \text{f}(x)$, we have $\text{f}(-2) = 1$:
$(-2)^3 + a(-2)^2 + b(-2) + c = 1$
$-8 + 4a - 2b + c = 1$
$4a - 2b + c = 9 \quad \text{--- (1)}$

Since $(2, -3)$ lies on $y = \text{f}(x)$, we have $\text{f}(2) = -3$:
$(2)^3 + a(2)^2 + b(2) + c = -3$
$8 + 4a + 2b + c = -3$
$4a + 2b + c = -11 \quad \text{--- (2)}$

2. Form an equation from the point on the translated graph $y = \text{f}(x + 1)$.

Since $(2, 1)$ lies on the graph of $y = \text{f}(x + 1)$, we substitute $x = 2$ and $y = 1$ into the equation:
$1 = \text{f}(2 + 1)$
$\text{f}(3) = 1$
$3^3 + a(3)^2 + b(3) + c = 1$
$27 + 9a + 3b + c = 1$
$9a + 3b + c = -26 \quad \text{--- (3)}$

3. Solve the system of 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 of equations can also be solved directly using the system solver function on a graphing calculator.)