How do you write (1-9i)(1-4i)(4-3i) in standard form?

In standard form $\left(1 - 9 i\right) \left(1 - 4 i\right) \left(4 - 3 i\right) = - 179 + 53 i$
$\left(1 - 9 i\right) \left(1 - 4 i\right) \left(4 - 3 i\right) = \left(1 - 13 i - 36\right) \left(4 - 3 i\right) \left[{i}^{2} = - 1\right]$
$\left(- 35 - 13 i\right) \left(4 - 3 i\right) = - 140 + 53 i - 39 = - 179 + 53 i$
In standard form $\left(1 - 9 i\right) \left(1 - 4 i\right) \left(4 - 3 i\right) = - 179 + 53 i$ [Ans]