Question

How do you expand ${(x-1)}^3$ using binomial expansion?

Answer 1

Expending using Binomial theorem :

Binomial Theorem is given by :

${(a+b)}^n=\sum _{k=0}^n{}^n{C}_k{\left(a\right)}^{n-k}{\left(b\right)}^k$

Here, $a=x$, $b=-1$, $n=3$

${(x-1)}^3=\sum _{k=0}^3{}^3{C}_k{\left(x\right)}^{3-k}{(-1)}^k$

${(x-1)}^3={}^{3}C_0{x}^3{(-1)}^0+{}^{3}C_1{x}^2{(-1)}^1+{}^{3}C_2{x}^1{(-1)}^2+{}^{3}C_3{x}^0{(-1)}^3$

${(x-1)}^3=\frac{3!}{3!0!}{x}^3-\frac{3!}{1!(3-1)!}{x}^2+\frac{3!}{2!(3-2)!}{x}^1-\frac{3!}{3!(3-3)!}$ $\left[0!=1\right]$

${(x-1)}^3=x^3-3x^2+3x-1$

Hence, binomial expansion of ${(x-1)}^3$ is $x^3-3x^2+3x-1$.