Solve the equation t4 + 4t3 − 6t2 − 4t + 1 = 0.

Since $t=0$ would lead to $1=0$, $t \neq 0$. We can safely divide by $t^2$.

Let $y = t - \frac{1}{t}$.
Then $y^2 = (t - \frac{1}{t})^2 = t^2 - 2(t)(\frac{1}{t}) + \frac{1}{t^2} = t^2 - 2 + \frac{1}{t^2}$.
So, $t^2 + \frac{1}{t^2} = y^2 + 2$.

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

Case 1: $y = -2 + 2\sqrt{2}$

Multiply by $t$:

Using the quadratic formula for $t$:

To simplify $\sqrt{12 - 8\sqrt{2}}$, we look for $(a-b\sqrt{2})^2 = a^2 + 2b^2 - 2ab\sqrt{2}$.
Comparing $12 - 8\sqrt{2}$ with $a^2+2b^2 - 2ab\sqrt{2}$, we have $2ab=8 \implies ab=4$ and $a^2+2b^2=12$.
If $a=2, b=2$, then $a^2+2b^2 = 4+2(4) = 12$. So $\sqrt{12 - 8\sqrt{2}} = \sqrt{(2-2\sqrt{2})^2} = |2-2\sqrt{2}| = 2\sqrt{2}-2$.

Two solutions from this case:

This solution is extraneous because we divided by $t^2$ earlier, assuming $t \neq 0$.

Case 2: $y = -2 - 2\sqrt{2}$

Multiply by $t$:

Using the quadratic formula for $t$:

To simplify $\sqrt{12 + 8\sqrt{2}}$, we look for $(a+b\sqrt{2})^2 = a^2 + 2b^2 + 2ab\sqrt{2}$.
Comparing $12 + 8\sqrt{2}$ with $a^2+2b^2 + 2ab\sqrt{2}$, we have $2ab=8 \implies ab=4$ and $a^2+2b^2=12$.
If $a=2, b=2$, then $a^2+2b^2 = 4+2(4) = 12$. So $\sqrt{12 + 8\sqrt{2}} = \sqrt{(2+2\sqrt{2})^2} = 2+2\sqrt{2}$.

Two solutions from this case:

This solution is also extraneous.

The valid solutions for $t$ are: