# How do you use the binomial formula to expand [x+(y+1)]^3?

Jun 15, 2016

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

#### Explanation:

This binomial has the form ${\left(a + b\right)}^{3}$
We expand the binomial by applying this property:
${\left(a + b\right)}^{3} = {a}^{3} + 3 {a}^{2} b + 3 a {b}^{2} + {b}^{3}$.

Where in given binomial $a = x$ and $b = y + 1$

We have:
${\left[x + \left(y + 1\right)\right]}^{3} =$
${x}^{3} + 3 {x}^{2} \left(y + 1\right) + 3 x {\left(y + 1\right)}^{2} + {\left(y + 1\right)}^{3}$ remark it as (1)

In the above expand we still have two binomials to expand
${\left(y + 1\right)}^{3}$ and ${\left(y + 1\right)}^{2}$

For ${\left(y + 1\right)}^{3}$ we have to use the above cubed property
So ${\left(y + 1\right)}^{3} = {y}^{3} + 3 {y}^{2} + 3 y + 1$. Remark it as (2)

For ${\left(y + 1\right)}^{2}$ we have to use the squared of the sum that says:
${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

So ${\left(y + 1\right)}^{2} = {y}^{2} + 2 y + 1$. Remark it as (3)

Substituting (2) and (3) in equation (1) we have:

${x}^{3} + 3 {x}^{2} \left(y + 1\right) + 3 x {\left(y + 1\right)}^{2} + {\left(y + 1\right)}^{3}$
$= {x}^{3} + 3 {x}^{2} \left(y + 1\right) + 3 x \left({y}^{2} + 2 y + 1\right) + \left({y}^{3} + 3 {y}^{2} + 3 y + 1\right)$
$= {x}^{3} + 3 {x}^{2} y + 3 {x}^{2} + 3 x {y}^{2} + 6 x y + 3 x + {y}^{3} + 3 {y}^{2} + 3 y + 1$
We have to add the similar terms but in this polynomial we don't have similar terms , we can arrange the terms .
Thus,

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