# Tom wrote 3 consecutive natural numbers. From these numbers' cube sum he took away the triple product of those numbers and divided by the arithmetic average of those numbers. What number did Tom write?

Oct 17, 2017

Final number that Tom wrote was $\textcolor{red}{9}$

#### Explanation:

Note: much of this depend upon my correctly understanding the meaning of various parts of the question.

3 consecutive natural numbers
I assume this could be represented by the set $\left\{\left(a - 1\right) , a , \left(a + 1\right)\right\}$ for some $a \in \mathbb{N}$

these numbers' cube sum
I assume this could be represented as
$\textcolor{w h i t e}{\text{XXX}} {\left(a - 1\right)}^{3} + {a}^{3} + {\left(a + 1\right)}^{3}$

$\textcolor{w h i t e}{\text{XXXXX}} = {a}^{3} - 3 {a}^{2} + 3 a - 1$
$\textcolor{w h i t e}{\text{XXXXXx}} + {a}^{3}$
$\textcolor{w h i t e}{\text{XXXXXx}} \underline{+ {a}^{3} + 3 {a}^{2} + 3 a + 1}$
$\textcolor{w h i t e}{\text{XXXXX}} = 3 {a}^{3} \textcolor{w h i t e}{+ 3 {a}^{2}} + 6 a$

the triple product of these numbers
I assume this means triple the product of these numbers
$\textcolor{w h i t e}{\text{XXX}} 3 \left(a - 1\right) a \left(a + 1\right)$
$\textcolor{w h i t e}{\text{XXXXX}} = 3 {a}^{3} - 3 a$

So these numbers' cube sum minus the triple product of these numbers would be
$\textcolor{w h i t e}{\text{XXXXX}} 3 {a}^{3} + 6 a$
$\textcolor{w h i t e}{\text{XXX}} \underline{- \left(3 {a}^{3} - 3 a\right)}$
$\textcolor{w h i t e}{\text{XXX")=color(white)("XXxX}} 9 a$

the arithmetic average of these three numbers
$\textcolor{w h i t e}{\text{XXX")((a-1)+a+(a+1))/3color(white)("XXX}} = a$

$\textcolor{w h i t e}{\text{XXX")(9a)/acolor(white)("XXX}} = 9$