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

Jun 30, 2016

In standard form $y = {x}^{3} - 9 {x}^{2} + 27 x - 27$

Explanation:

In $y = {\left(x - 3\right)}^{3}$, the RHS is a polynomial of degree $3$ in $x$.

The standard form of a polynomial in degree $3$ is $a {x}^{3} + b {x}^{2} + c x + d$, so we should expend ${\left(x - 3\right)}^{3}$ by multiplying.

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

= $\left(x - 3\right) \left(x \left(x - 3\right) - 3 \left(x - 3\right)\right)$

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

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

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

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

= ${x}^{3} - 9 {x}^{2} + 27 x - 27$