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

Aug 4, 2015

Make the problem symmetrical by evaluating ${\left(x + y + z\right)}^{3}$ then substitute $z = 1$ to find:

${\left(x + \left(y + 1\right)\right)}^{3} = {x}^{3} + {y}^{3} + 1 + 3 {x}^{2} y + 3 {y}^{2} + 3 x + 3 {x}^{2} + 3 {y}^{2} x + 3 y + 6 x y$

#### Explanation:

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

The coefficient of ${x}^{3}$ in the product is the number of ways of choosing a term from each of the trinomials so their product is ${x}^{3}$. This can only be done one way, by choosing $x$ from each of the three trinomials. So the coefficient of ${x}^{3}$ is $1$. By symmetry, the coefficients of ${y}^{3}$ and ${z}^{3}$ are also $1$.

The coefficient of ${x}^{2} y$ in the product is the number of ways of choosing a term from each of the trinomials so their product is ${x}^{2} y$. We can choose one of the $3$ trinomials to take $y$ from, then choose $x$ from the other two trinomials. This can be done in $3$ ways, so the coefficient of ${x}^{2} y$ in the product is $3$. By symmetry, $3$ is also the coefficient of ${y}^{2} z$, ${z}^{2} x$, ${x}^{2} z$, ${y}^{2} x$ and $z {x}^{2}$.

The coefficient of $x y z$ in the product is the number of ways of choosing a term from each of the trinomials so their product is $x y z$. There are $3$ ways to choose a trinomial to take $x$ from. Then there are $2$ remaining trinomials to pick $y$ from. That leaves one trinomial to take $z$ from. So the number of ways of forming $x y z$ is $3 \times 2 \times 1 = 6$.

So

${\left(x + y + z\right)}^{3} = {x}^{3} + {y}^{3} + {z}^{3} + 3 {x}^{2} y + 3 {y}^{2} z + 3 {z}^{2} x + 3 {x}^{2} z + 3 {y}^{2} x + 3 {z}^{2} y + 6 x y z$

Now let $z = 1$ to get:

${\left(x + \left(y + 1\right)\right)}^{3} = {x}^{3} + {y}^{3} + 1 + 3 {x}^{2} y + 3 {y}^{2} + 3 x + 3 {x}^{2} + 3 {y}^{2} x + 3 y + 6 x y$