Show that 1 is a solution of the equation $x^3 - 5x^2 = 2 - 6x$. Hence solve the equation $x^3 - 5x^2 = 2 - 6x$ completely, expressing non-integer roots in surd form. Hence solve the equation $(x-1)^3 - 5(x-1)^2 + 6x - 8 = 0$ completely. Express non-integer roots in surd form.

1. Rearrange the equation and show that $x = 1$ is a solution

Rearranging the equation gives $x^3 - 5x^2 + 6x - 2 = 0$.
Let $P(x) = x^3 - 5x^2 + 6x - 2$.
Substitute $x = 1$:
$P(1) = (1)^3 - 5(1)^2 + 6(1) - 2$
$P(1) = 1 - 5 + 6 - 2 = 0$
Since $P(1) = 0$, $x = 1$ is a solution to the equation.

2. Find the quadratic factor of the cubic polynomial

Since $x = 1$ is a root, $(x - 1)$ is a factor.
$x^3 - 5x^2 + 6x - 2 = (x - 1)(x^2 + bx + c)$
By comparing coefficients:
Constant term: $-1 \times c = -2 \implies c = 2$
Coefficient of $x^2$: $-1 + b = -5 \implies b = -4$
Thus, the quadratic factor is $x^2 - 4x + 2$.

3. Solve the quadratic equation to find the remaining roots

Set $x^2 - 4x + 2 = 0$.
Using the quadratic formula:
$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}$
$x = \frac{4 \pm \sqrt{16 - 8}}{2}$
$x = \frac{4 \pm \sqrt{8}}{2}$

4. Simplify the surd to obtain the final roots

$x = \frac{4 \pm 2\sqrt{2}}{2}$
$x = 2 \pm \sqrt{2}$
The solutions are $x = 1$, $x = 2 + \sqrt{2}$, and $x = 2 - \sqrt{2}$.

1. Rewrite the given equation to match the form of the equation in part (a).

The equation is $(x-1)^3 - 5(x-1)^2 + 6x - 8 = 0$.
Notice that $6x - 8 = 6(x-1) - 2$.
So the equation can be written as:
$(x-1)^3 - 5(x-1)^2 + 6(x-1) - 2 = 0$

2. Use the roots from part (a) to find the solutions for $x$.

From part (a), the roots of $y^3 - 5y^2 + 6y - 2 = 0$ are $y = 1$ and $y = 2 \pm \sqrt{2}$.
Let $y = x - 1$.
Then $x - 1 = 1 \implies x = 2$
$x - 1 = 2 + \sqrt{2} \implies x = 3 + \sqrt{2}$
$x - 1 = 2 - \sqrt{2} \implies x = 3 - \sqrt{2}$