The cubic polynomial $\text{f}(x)$ is such that the coefficient of $x^3$ is $-1$ and the roots of $\text{f}(x) = 0$ are $1$, $k$ and $k^2$. It is given that $\text{f}(x)$ has a remainder of $-7$ when divided by $x - 2$.

1. Formulate the expression for the cubic polynomial $f(x)$ using its roots and leading coefficient.

Since the roots are $1$, $k$, and $k^2$, and the coefficient of $x^3$ is $-1$, we can write:
$f(x) = -1(x - 1)(x - k)(x - k^2)$

2. Apply the Remainder Theorem to set up an equation.

By the Remainder Theorem, since $f(x)$ has a remainder of $-7$ when divided by $x - 2$, we have $f(2) = -7$.
$-1(2 - 1)(2 - k)(2 - k^2) = -7$

3. Expand the equation and simplify to obtain the required result.

$-1(1)(4 - 2k^2 - 2k + k^3) = -7$
$4 - 2k^2 - 2k + k^3 = 7$
$k^3 - 2k^2 - 2k - 3 = 0 \quad \text{(shown)}$

1. Find an integer root of the cubic equation using the Factor Theorem

Let $P(k) = k^3 - 2k^2 - 2k - 3$.
Try $k = 3$:
$P(3) = (3)^3 - 2(3)^2 - 2(3) - 3 = 27 - 18 - 6 - 3 = 0$
So $k = 3$ is a root, and $(k - 3)$ is a factor.

2. Factorise the cubic polynomial completely

$k^3 - 2k^2 - 2k - 3 = (k - 3)(k^2 + ak + 1)$
Comparing coefficients of $k^2$:
$-3 + a = -2 \implies a = 1$
So, $k^3 - 2k^2 - 2k - 3 = (k - 3)(k^2 + k + 1) = 0$

3. State the real value of k

From the linear factor, $k - 3 = 0 \implies k = 3$.

4. Show that the quadratic factor has no real roots

For the quadratic factor $k^2 + k + 1 = 0$, we check the discriminant:
$\Delta = b^2 - 4ac = (1)^2 - 4(1)(1) = 1 - 4 = -3$
Since $\Delta < 0$, the equation $k^2 + k + 1 = 0$ has no real roots. Therefore, $k = 3$ is the only real value.