Question

Factor:$432c^3+686d^3$.

Short answer

Factors of$432c^3+686d^3$are$2\left(6c+7d\right)\left(36c^2-42cd+49d^2\right)$.

Step-by-step solution

Step 1. Given information.

Given an expression $432c^3+686d^3$.

Step 2. Factor the common factor.

Common factor of the given expression is 2.

So, the expression can be expressed as$2\left(216c^3+343d^3\right)$.

Step 3. Check if the binomial 216c3+343d3 is a sum of perfect cubes.

Consider $a=6c$and $b=7d$.

It can be seen that $216c^3+343d^3={\left(6c\right)}^3+{\left(7d\right)}^3$.

It follows that the expression $216c^3+343d^3$is sum of perfect cubes.

Step 4. Use the sum of cubes pattern and simplify.

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

Obtain the result as follows.

$\begin{array}{rcl}2\left(216c^3+343d^3\right)& =& 2\left({\left(6c\right)}^3+{\left(7d\right)}^3\right)\\ & =& 2\left(6c+7d\right)\left({\left(6c\right)}^2-42cd+{\left(7d\right)}^2\right)\\ & =& 2\left(6c+7d\right)\left(36c^2-42cd+49d^2\right)\end{array}$

Step 5. Check the result by multiplying the factors.

Verify the result as follows.

$\begin{array}{rcl}2\left(6c+7d\right)\left(36c^2-42cd+49d^2\right)& =& 2\left(6c\left(36c^2-42cd+49d^2\right)+7d\left(36c^2-42cd+49d^2\right)\right)\\ & =& 2\left(216c^3-252c^{2}d+294c{d}^2+252d{c}^2-294c{d}^2+343d^3\right)\\ & =& 2\left(216c^3+343d^3\right)\\ & =& 432c^3+686d^3\end{array}$