Question

How do you factor ${{x}^{6}}-8{{y}^{3}}$?

Answer 1

Hint: To factor the given polynomial ${{x}^{6}}-8{{y}^{3}}$, we have to use the algebraic identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. For this, we need to write the given polynomial in the form of ${{a}^{3}}-{{b}^{3}}$. And for this, we have to put $8={{2}^{3}}$ in the given polynomial ${{x}^{6}}-8{{y}^{3}}$. Then we can put ${{x}^{6}}={{\left( {{x}^{2}} \right)}^{3}}$ and $8{{y}^{3}}={{\left( 2y \right)}^{3}}$ so that the given polynomial will become ${{\left( {{x}^{2}} \right)}^{3}}-{{\left( 2y \right)}^{3}}$. Then on putting $a={{x}^{2}}$ and $b=2y$ in the algebraic identity ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$, the given polynomial will be completely factored.

Complete step by step solution:
Let us consider the polynomial given in the above question as
$\Rightarrow p\left( x,y \right)={{x}^{6}}-8{{y}^{3}}$
Putting $8={{2}^{3}}$ and ${{x}^{6}}={{\left( {{x}^{2}} \right)}^{3}}$ in the above polynomial, we get
$\Rightarrow p\left( x,y \right)={{\left( {{x}^{2}} \right)}^{3}}-{{2}^{3}}{{y}^{3}}$
Using the exponent property given by ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, we can write the above polynomial as
$\Rightarrow p\left( x,y \right)={{\left( {{x}^{2}} \right)}^{3}}-{{\left( 2y \right)}^{3}}$
So we have expressed the given polynomial as a difference of two cubes. The algebraic identity of the difference of cubes is given by ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. Using this algebraic identity we can write the above polynomial as
\[\begin{align}
  & \Rightarrow p\left( x,y \right)=\left( {{x}^{2}}-2y \right)\left( {{\left( {{x}^{2}} \right)}^{2}}+2{{x}^{2}}y+{{\left( 2y \right)}^{2}} \right) \\
 & \Rightarrow p\left( x,y \right)=\left( {{x}^{2}}-2y \right)\left( {{x}^{4}}+2{{x}^{2}}y+4{{y}^{2}} \right) \\
\end{align}\]
Hence, the given polynomial is factored completely.

Note: Do not try to use the hit and trial method for factoring the given polynomial. This is because the given polynomial is not in terms of a single variable, but it is in terms of two variables, $x$ and $y$. So we are left with only the application of the algebraic identities to factor the given polynomial. Do not write $2y={{\left( \sqrt{2y} \right)}^{2}}$ in the factor \[\left( {{x}^{2}}-2y \right)\] since we do not know whether $y$ is positive or negative. Also, do not make the mistake of factoring the factor \[\left( {{\left( {{x}^{2}} \right)}^{2}}+2{{x}^{2}}y+{{\left( 2y \right)}^{2}} \right)\] using the identity ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$. This is because the middle term in the factor \[\left( {{\left( {{x}^{2}} \right)}^{2}}+2{{x}^{2}}y+{{\left( 2y \right)}^{2}} \right)\] for this must be $4{{x}^{2}}y$ and not $2{{x}^{2}}y$.