Q 6.68 - Intermediate Algebra

Question

Factor: ${\left(n+3\right)}^3-125n^3$.

Step-by-Step Solution

Answer

Factors of ${\left(n+3\right)}^3-125n^3$ are $\left(-4n+3\right)\left(31n^2+21n+9\right)$.

Step 2. Given information.

Given an expression ${\left(n+3\right)}^3-125n^3$.

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

It can be seen that the binomial is a difference.

Consider $a=n+3$ and $b=5n$.

Then, ${\left(n+1\right)}^3-125n^3={\left(n+1\right)}^3-{\left(5n\right)}^3$.

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.

$\begin{array}{rcl}{\left(n+3\right)}^3-125n^3& =& {\left(n+3\right)}^3-{\left(5n\right)}^3\\ & =& \left(n+3-5n\right)\left({\left(n+3\right)}^2+5n\left(n+3\right)+{\left(5n\right)}^2\right)\\ & =& \left(-4n+3\right)\left(n^2+6n+9+5n^2+15n+25n^2\right)\\ & =& \left(-4n+3\right)\left(31n^2+21n+9\right)\end{array}$

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.

$\begin{array}{rcl}\left(-4n+3\right)\left(31n^2+21n+9\right)& =& -4n\left(31n^2+21n+9\right)+3\left(31n^2+21n+9\right)\\ & =& -124n^3+9n^2+27n+27\\ & =& n^3+3^3+3{\left(n\right)}^{2}3+3n{\left(3\right)}^2-125n^3\\ & =& {\left(n+3\right)}^3-125n^3\end{array}$