Question

Factor:${\left(y+1\right)}^3-27y^3$.

Short answer

Factors of${\left(y+1\right)}^3-27y^3$are$\left(-2y+1\right)\left(13y^2+5y+1\right)$.

Step-by-step solution

Step 1. Given information.

Given an expression${\left(y+1\right)}^3-27y^3$.

Step 2. Check if the binomial is a difference of perfect cubes.

It can be seen that the binomial is a difference.

Consider $a=y+1$and $b=3y$.

Then, ${\left(y+1\right)}^3-27y^3={\left(y+1\right)}^3-{\left(3y\right)}^3$.

Therefore, the binomial is a difference of perfect cubes.

Step 3. Use the difference of cubes pattern and simplify.

Note that $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$.

Factor the expression and simplify as follows.

$\begin{array}{rcl}{\left(y+1\right)}^3-27y^3& =& {\left(y+1\right)}^3-{\left(3y\right)}^3\\ & =& \left(y+1-3y\right)\left({\left(y+1\right)}^2+3y\left(y+1\right)+{\left(3y\right)}^2\right)\\ & =& \left(-2y+1\right)\left(y^2+2y+1+3y^2+3y+9y^2\right)\\ & =& \left(-2y+1\right)\left(13y^2+5y+1\right)\end{array}$

Step 4. Check the result by multiplying the factors.

Note that ${\left(a+b\right)}^3=a^3+b^3+3a^{2}b+3a{b}^2$.

Verify the result as follows.

$\begin{array}{rcl}\left(-2y+1\right)\left(13y^2+5y+1\right)& =& -2y\left(13y^2+5y+1\right)+1\left(13y^2+5y+1\right)\\ & =& -26y^3+3y^2+3y+1\\ & =& y^3+1^3+3y^2\left(1\right)+3y{\left(1\right)}^2-27y^3\\ & =& {\left(y+1\right)}^3-27y^3\end{array}$