What is the standard form of  y= (x-1)^3-(x-3)^2?

Nov 28, 2015

$y = {x}^{3} - 4 {x}^{2} + 9 x - 10$

Explanation:

Calculate each part separately:

${\left(x - 1\right)}^{3} = {\overbrace{\left(x - 1\right) \left(x - 1\right)}}^{\text{multiply just these first}} \left(x - 1\right)$
$= \left({x}^{2} - x - x + 1\right) \left(x - 1\right)$
$= \left({x}^{2} - 2 x + 1\right) \left(x - 1\right)$
$= {x}^{3} - {x}^{2} - 2 {x}^{2} + 2 x + x - 1$
$= {x}^{3} - 3 {x}^{2} + 3 x - 1$

${\left(x - 3\right)}^{2} = \left(x - 3\right) \left(x - 3\right)$
$= {x}^{2} - 3 x - 3 x + 9$
$= {x}^{2} - 6 x + 9$

Put back into the original expression.

$y = \left({x}^{3} - 3 {x}^{2} + 3 x - 1\right) - \left({x}^{2} - 6 x + 9\right)$
$= \textcolor{red}{{x}^{3}} \textcolor{b l u e}{- 3 {x}^{2}} \textcolor{g r e e n}{+ 3 x} \textcolor{p u r p \le}{- 1} \textcolor{b l u e}{- {x}^{2}} \textcolor{g r e e n}{+ 6 x} \textcolor{p u r p \le}{- 9}$
$= {x}^{3} - 4 {x}^{2} + 9 x - 10$