Question

How do you factor ${{x}^{3}}+729$?

Answer 1

Hint: We first take the factorisation of the given polynomial ${{x}^{3}}+729$ according to the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. We form the factorisation to find the simplified form of ${{x}^{3}}+729$ by replacing with $a=x;b=9$. We also verify the result with an arbitrary value of $x$.

Complete step by step answer:
The given polynomial ${{x}^{3}}+729$ is cubic expression. We consider ${{x}^{3}}$ as ${{\left( x \right)}^{3}}$ and 729 as ${{9}^{3}}$.
It’s a sum of two cube numbers. We factorise the given sum of the cubes according to the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$.
We have ${{x}^{3}}+729$ and for the theorem we replace the values as $a=x;b=9$
We get \[{{x}^{3}}+729={{\left( x \right)}^{3}}+{{9}^{3}}=\left( x+9 \right)\left[ {{x}^{2}}-9x+81 \right]\].
We can see the term ${{x}^{3}}+729$ is a multiplication of two polynomials \[\left( x+9 \right)\] and \[\left( {{x}^{2}}-9x+81 \right)\].
These terms can’t be factored any more.
The factorisation of ${{x}^{3}}+729$ is \[\left( x+9 \right)\left( {{x}^{2}}-9x+81 \right)\].
Now we verify the result with an arbitrary value of $x=2$.
We have ${{x}^{3}}+729=\left( x+9 \right)\left( {{x}^{2}}-9x+81 \right)$.
The left-hand side of the equation gives ${{x}^{3}}+729={{2}^{3}}+729=8+729=737$.
The left-hand side of the equation gives
$\begin{align}
  & \left( x+9 \right)\left( {{x}^{2}}-9x+81 \right) \\
 & =\left( 2+9 \right)\left( {{2}^{2}}-9\times 2+81 \right) \\
 & =11\times 67 \\
 & =737 \\
\end{align}$
Thus, verified the result of ${{x}^{3}}+729=\left( x+9 \right)\left( {{x}^{2}}-9x+81 \right)$.

Note: We explain the process of getting ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$.
We need to find the simplified form of ${{\left( a+b \right)}^{3}}$. This is the cube of sum of two numbers.
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
We need to multiply the term $\left( a+b \right)$ on both side of the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.
On the left side of the equation, we get ${{\left( a+b \right)}^{2}}\left( a+b \right)={{\left( a+b \right)}^{3}}$.
On the right side we have $\left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right)$. We use multiplication and get
$\begin{align}
  & \Rightarrow \left( {{a}^{2}}+{{b}^{2}}+2ab \right)\left( a+b \right) \\
 & ={{a}^{2}}.a+a.{{b}^{2}}+2ab\times a+{{a}^{2}}.b+{{b}^{2}}.b+2ab.b \\
 & ={{a}^{3}}+a{{b}^{2}}+2{{a}^{2}}b+{{a}^{2}}b+{{b}^{3}}+2a{{b}^{2}} \\
 & ={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}} \\
\end{align}$
We also can take another form where
${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$.
This gives
$\begin{align}
  & {{a}^{3}}+{{b}^{3}} \\
 & ={{\left( a+b \right)}^{3}}-3ab\left( a+b \right) \\
 & =\left( a+b \right)\left[ {{\left( a+b \right)}^{2}}-3ab \right] \\
 & =\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
\end{align}$