$x^4 - 4x^3 + 6x^2 - 4x + 1 = 0$

Two ways:

First, look at the coefficients: $1, 4, 6, 4, 1$ with alternating signs. If you know the binomial theorem, you can easily associate this to the expansion of $(x-1)^4$. You'd have to be pretty darn clever.

Second, you can just find the roots via synthetic division. You'll see that you'll have $x=1$ as the only root with multiplicity $4$. This corresponds to the factors $(x-1)(x-1)(x-1)(x-1)$.

You can even check that $(x-1)^4$ yields the above expansion by multiplying.

You can factor it as

$$x^{ 4 }-4x^{ 3 }+6x^{ 2 }-4x+1={ x }^{ 4 }-2{ x }^{ 3 }+{ x }^{ 2 }-2{ x }^{ 3 }+4{ x }^{ 2 }-2x+{ x }^{ 2 }-2x+1={ x }^{ 2 }\left( { x }^{ 2 }-2x+1 \right) -2x\left( { x }^{ 2 }-2x+1 \right) +{ x }^{ 2 }-2x+1=\\ =\left( { x }^{ 2 }-2x+1 \right) \left( { x }^{ 2 }-2x+1 \right) ={ \left( x-1 \right) }^{ 2 }{ \left( x-1 \right) }^{ 2 }={ \left( x-1 \right) }^{ 4 }\\ $$

The rational root theorem tells you that any rational root $\frac{p}{q}$ will have $p\mid 1$ and $q \mid 1$ so $\frac{p}{q} = \pm 1$, you can then verify that only $x=1$ is a root and be done.