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

Jan 12, 2016

${x}^{3} - 10 {x}^{2} + 21 x - 36$

#### Explanation:

To obtain standard form require to expand brackets and collect like terms.

${\left(x - 3\right)}^{3} - {\left(x + 3\right)}^{2}$ can be rewritten as follows :

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

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

now becomes ;

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

expanding both pairs of brackets :

${x}^{3} - 6 {x}^{2} + 9 x - 3 {x}^{2} + 18 x - 27 - \left({x}^{2} + 6 x + 9\right)$

now rewriting with no brackets :

${x}^{3} - 6 {x}^{2} + 9 x - 3 {x}^{2} + 18 x - 27 - {x}^{2} - 6 x - 9$

Finally collect like terms and write expression in descending order ie. term with highest power → term with lowest power (usually constant term.

$\Rightarrow {\left(x - 3\right)}^{3} - {\left(x + 3\right)}^{2} = {x}^{3} - 10 {x}^{2} + 21 x - 36$