Expand ${(x-1)}^4$.

Use the binomial theorem

We know that the binomial expansion of ${(x-a)}^n$ is given by ${}^n{C}_0{x}^n+{}^{n}C_1{x}^{n-1}{a}^1+{}^n{C}_2{x}^{n-2}{a}^2+{}^n{C}_3{x}^{n-3}{a}^3+....+{}^n{C}_n{x}^0{a}^n$

So using the above formula we will expand ${(x-1)}^4$

Here,$a=x$ and $b=-1.$

$\Rightarrow {\left(x-1\right)}^4={}^4{C}_0{x}^4+{}^{4}C_1{x}^3\times {(-1)}^1+{}^4{C}_2{x}^2\times {(-1)}^2+{}^4{C}_{3}x{(-1)}^3+{}^4{C}_4{(-1)}^4.......\left(1\right)$

We know that ${}^4{C}_0={}^4{C}_4=1,{}^4{C}_1={}^4{C}_3=4$ and ${}^{4}C_2=6$

On substituting above values in the above equation $\left(1\right)$, we get

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

Hence, the expansion of ${(x-1)}^4$ is $x^4-4x^3+6x^2-4x+1$