How do you multiply (x-3)^5?

${\left(x - 3\right)}^{5} = {x}^{5} - 15 {x}^{4} + 90 {x}^{3} - 270 {x}^{2} + 405 x - 243$
Either use binomial theorem to get the answer or using the Pascal's triangle, the expansion of ${\left(x + y\right)}^{5} = {x}^{5} + 5 {x}^{4} y + 10 {x}^{3} {y}^{2} + 10 {x}^{2} {y}^{3} + 5 x {y}^{4} + {y}^{5}$
${\left(x - 3\right)}^{5} = {x}^{5} - 15 {x}^{4} + 90 {x}^{3} - 270 {x}^{2} + 405 x - 243$