# How do you multiply (2-x)^3?

${\left(2 - x\right)}^{3} = 8 - 12 x + 6 {x}^{2} - {x}^{3}$

#### Explanation:

To multiply ${\left(2 - x\right)}^{3}$, we have several ways to do it. One solution is by Binomial Theorem and another is by simply multiplying the expression $\left(2 - x\right)$ by itself and the result by itself again.

Solution by Binomial Theorem

${\left(a - b\right)}^{3} = {a}^{3} - 3 {a}^{2} b + 3 a {b}^{2} - {b}^{3}$

So that

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

${\left(2 - x\right)}^{3} = 8 - 12 x + 6 {x}^{2} - {x}^{3}$

Solution by multiplication

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

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

${\left(2 - x\right)}^{3} = \left[4 - 2 x - 2 x + {x}^{2}\right] \cdot \left(2 - x\right)$

${\left(2 - x\right)}^{3} = \left[4 - 4 x + {x}^{2}\right] \cdot \left(2 - x\right)$

${\left(2 - x\right)}^{3} = \left[4 \cdot \left(2 - x\right) - 4 x \cdot \left(2 - x\right) + {x}^{2} \cdot \left(2 - x\right)\right]$

${\left(2 - x\right)}^{3} = \left[8 - 4 x - 8 x + 4 {x}^{2} + 2 {x}^{2} - {x}^{3}\right]$

${\left(2 - x\right)}^{3} = 8 - 12 x + 6 {x}^{2} - {x}^{3}$

God bless....I hope the explanation is useful.