To determine when the particle changes direction, you need to find when the velocity changes sign. The velocity is the derivative of the position function.

Given the position function:

$x(t) = t^3 - 9t^2 - 21t + 6$

First, find the velocity $v(t)$ by differentiating $x(t)$:

$v(t) = \frac{d}{dt}(t^3 - 9t^2 - 21t + 6) = 3t^2 - 18t - 21$

To find when the particle changes direction, set $v(t) = 0$ and solve for $t$:

$3t^2 - 18t - 21 = 0$

Divide the equation by 3:

$t^2 - 6t - 7 = 0$

Solve this quadratic equation using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -6$, and $c = -7$.

$t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1}$

$t = \frac{6 \pm \sqrt{36 + 28}}{2}$

$t = \frac{6 \pm \sqrt{64}}{2}$

$t = \frac{6 \pm 8}{2}$

So, $t = 7$ or $t = -1$.

Since $t \geq 0$, the particle changes direction at $t = 7$.