# What is the standard form of  y= (4x-4)(6x-2)(8x-4)?

$y = 96 {x}^{3} - 128 x + 32$
$= \left(24 {x}^{2} - 32 x + 8\right) \cdot \left(8 x - 4\right)$
$= \left(192 {x}^{3} - 256 x + 64 x - 96 {x}^{3} + 128 x - 32\right)$
$= \left(96 {x}^{3} - 128 x + 32\right)$