Question

How do you factor ${{x}^{6}}+8$?

Answer 1

Hint: We first take the factorisation of the given polynomial ${{x}^{6}}+8$ according to the identity ${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)$. We solve the multiplication to find the simplified form of ${{\left( y-4 \right)}^{3}}$ by replacing with $a=y;b=4$. We also verify the result with an arbitrary value of x.

Complete step by step answer:
The given polynomial ${{x}^{6}}+8$ is cubic expression. We consider ${{x}^{6}}$ as ${{\left( {{x}^{2}} \right)}^{3}}$ and 8 as ${{2}^{3}}$.
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 ${{x}^{6}}+8$ and for the theorem we replace the values as $a={{x}^{2}};b=2$
We get \[{{x}^{6}}+8={{\left( {{x}^{2}} \right)}^{3}}+{{2}^{3}}=\left( {{x}^{2}}+2 \right)\left[ {{x}^{4}}-2{{x}^{2}}+4 \right]\].
We can see the term ${{x}^{6}}+8$ is a multiplication of two polynomials \[\left( {{x}^{2}}+2 \right)\] and \[\left( {{x}^{4}}-2{{x}^{2}}+4 \right)\].
These terms can’t be factored any more.
The factorisation of ${{x}^{6}}+8$ is \[\left( {{x}^{2}}+2 \right)\left( {{x}^{4}}-2{{x}^{2}}+4 \right)\].
Now we verify the result with an arbitrary value of $x=2$.
We have ${{x}^{6}}+8=\left( {{x}^{2}}+2 \right)\left( {{x}^{4}}-2{{x}^{2}}+4 \right)$.
The left-hand side of the equation gives ${{x}^{6}}+8={{2}^{6}}+8=64+8=72$.
The left-hand side of the equation gives
$\begin{align}
  & \left( {{x}^{2}}+2 \right)\left( {{x}^{4}}-2{{x}^{2}}+4 \right) \\
 & =\left( {{2}^{2}}+2 \right)\left( {{2}^{4}}-2\times {{2}^{2}}+4 \right) \\
 & =6\times 12 \\
 & =72 \\
\end{align}$
Thus, verified the result of ${{x}^{6}}+8=\left( {{x}^{2}}+2 \right)\left( {{x}^{4}}-2{{x}^{2}}+4 \right)$.

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 the 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
$\begin{align}
  & {{a}^{3}}+{{b}^{3}} \\
 & ={{\left( a+b \right)}^{3}}-3ab\left( a+b \right) \\
 & =\left( a+b \right)\left[ {{\left( a+b \right)}^{2}}-3ab \right] \\
 & =\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) \\
\end{align}$