Question

How do you factor $27{{x}^{3}}-512$?

Answer 1

Hint: We can observe that the given equation contains one cubic variable in subtraction. In algebra we have the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. But in the given equation we have additionally coefficient for the variable and a constant, so we need to factorize the coefficient of the given variable, constant at the same time and we will convert the given equation in form of ${{a}^{3}}+{{b}^{3}}$, by applying some exponential rules, then we will use the algebraic formula and write the given equation as the product of its factors.

Complete step by step solution:
Given equation $27{{x}^{3}}-512$.
Considering the coefficient of ${{x}^{3}}$, which is $27$. Factorizing $27$.
Dividing $27$ with $2$. We can’t divide the $27$ with $2$. So, checking whether the value $27$ is divided by $3$. When we divide the number $27$ with $3$, then we will get
$27=3\times 9$.
Now we can write the number $9$ as $3\times 3$. From this we can write the number $27$ as
$\Rightarrow 27=3\times 3\times 3$
We have an exponential rule $a\times a\times a\times ....n\text{ times}={{a}^{n}}$, then we will have
$\therefore 27={{3}^{3}}$
Considering the constant which is $512$. Factorizing $512$.
Dividing $512$ with $2$. We will get $256$ as quotient, so we can write
$512=2\times 256$
Again dividing $256$ with $2$. We will get $128$ as quotient, so we can write
$256=2\times 128$
Again dividing $128$ with $2$. We will get $64$ as quotient, so we can write
$128=2\times 64$
Again dividing $64$ with $2$. We will get $32$ as quotient, so we can write
$64=2\times 32$
Again dividing $32$ with $2$. We will get $16$ as quotient, so we can write
$32=2\times 16$
Again dividing $16$ with $2$. We will get $8$ as quotient, so we can write
$16=2\times 8$
We can write the value $8$ as $2\times 2\times 2$. From the above values we are going to write the term $512$ as
$\begin{align}
  & \Rightarrow 512=2\times 256 \\
 & \Rightarrow 512=2\times 2\times 128 \\
 & \Rightarrow 512=2\times 2\times 2\times 64 \\
 & \Rightarrow 512=2\times 2\times 2\times 2\times 32 \\
 & \Rightarrow 512=2\times 2\times 2\times 2\times 2\times 16 \\
 & \Rightarrow 512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2 \\
\end{align}$
Simplifying the above equation, then we will get
$\begin{align}
  & \Rightarrow 512=\left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right)\times \left( 2\times 2\times 2 \right) \\
 & \Rightarrow 512=8\times 8\times 8 \\
\end{align}$
We have an exponential rule $a\times a\times a\times ....n\text{ times}={{a}^{n}}$, then we will have
$\therefore 512={{8}^{3}}$
Now the given equation is modified as
$\Rightarrow 27{{x}^{3}}-512={{3}^{3}}{{x}^{3}}-{{8}^{3}}$
We have the exponential rule ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, then we will get
$\Rightarrow 27{{x}^{3}}-512={{\left( 3x \right)}^{3}}-{{8}^{3}}$
Applying the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow 27{{x}^{3}}-512=\left( 3x-8 \right)\left[ {{\left( 3x \right)}^{2}}+\left( 3x \right)\left( 8 \right)+{{8}^{2}} \right] \\
 & \Rightarrow 27{{x}^{3}}-512=\left( 3x-8 \right)\left( 9{{x}^{2}}+24x+64 \right) \\
\end{align}$

Note: For this question we have given the difference of two cubes, so we have calculated the factors by using the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. If they have given the sum of the cubes, then we will use the formula ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$ and simplify the equation to get the required result.