How do you cube  [x+(y+1)]^3?

Aug 7, 2015

Solve the general case of ${\left(x + y + z\right)}^{3}$, then substitute $z = 1$ to find:

${\left(x + y + 1\right)}^{3}$

$= {x}^{3} + {y}^{3} + 1 + 3 {x}^{2} y + 3 {y}^{2} + 3 x + 3 {x}^{2} + 3 {y}^{2} x + 3 y + 6 x y$

Explanation:

The general case of ${\left(x + y + z\right)}^{3}$ is easier to handle due to its symmetry in $x$, $y$ and $z$. So solve that first, then substitute $z = 1$ ...

${\left(x + y + z\right)}^{3} = \left(x + y + z\right) \left(x + y + z\right) \left(x + y + z\right)$

The coefficient of the resulting ${x}^{3}$ term will be $1$, since there's only one way of picking $x$ from each of the $3$ trinomials. By symmetry the coefficient of ${y}^{3}$ and ${z}^{3}$ is also $1$.

The coefficient of the resulting ${x}^{2} y$ term will be $3$, since there are $3$ trinomials to pick $y$ from, then one way to pick $x$ from both of the remaining trinomials. By symmetry, $3$ is also the coefficient of ${y}^{2} z$, ${z}^{2} x$, ${x}^{2} z$, ${y}^{2} x$ and ${z}^{2} y$.

The coefficient of the resulting $x y z$ term will be $6$, since therre are $3$ trinomials to pick $x$ from, $2$ remaining trinomials to pick $y$ from, then only one remaining trinomial to pick $z$ from.

Putting this all together:

${\left(x + y + z\right)}^{3}$

$= {x}^{3} + {y}^{3} + {z}^{3} + 3 {x}^{2} y + 3 {y}^{2} z + 3 {z}^{2} x + 3 {x}^{2} z + 3 {y}^{2} x + 3 {z}^{2} y + 6 x y z$

Now substitute $z = 1$ to get:

${\left(x + y + 1\right)}^{3}$

$= {x}^{3} + {y}^{3} + 1 + 3 {x}^{2} y + 3 {y}^{2} + 3 x + 3 {x}^{2} + 3 {y}^{2} x + 3 y + 6 x y$