Question

How do you factor \[{{x}^{6}}{{y}^{3}}+{{y}^{9}}\]?

Answer 1

Hint: Take \[{{y}^{3}}\] common from both the terms and write the remaining terms in the bracket. Now, compare the terms inside the bracket with the algebraic expression \[{{a}^{3}}+{{b}^{3}}\] by writing \[{{x}^{6}}+{{y}^{6}}\] as \[{{\left( {{x}^{2}} \right)}^{3}}+{{\left( {{y}^{2}} \right)}^{3}}\]. Use the algebraic identity: - \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\] to simplify and get the answer.

Complete step by step answer:
Here, we have been provided with the expression \[{{x}^{6}}{{y}^{3}}+{{y}^{9}}\] and we are asked to factorize it.
Now, taking \[{{y}^{3}}\] common from the two terms, we have,
\[\Rightarrow {{x}^{6}}{{y}^{3}}+{{y}^{9}}={{y}^{3}}\left( {{x}^{6}}+{{y}^{6}} \right)\]
We have a property of ‘exponents and powers’ given as: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], so we can write \[{{x}^{6}}\] as \[{{x}^{2\times 3}}\] and it can be further simplified as \[{{\left( {{x}^{2}} \right)}^{3}}\]. Using similar property for \[{{y}^{2}}\], we can write the expression as: -
\[\Rightarrow {{x}^{6}}{{y}^{3}}+{{y}^{9}}={{y}^{3}}\left[ {{\left( {{x}^{2}} \right)}^{3}}+{{\left( {{y}^{2}} \right)}^{3}} \right]\]
In the R.H.S. of the above expression we have an expression similar to the algebraic expression \[{{a}^{3}}+{{b}^{3}}\] whose simplified or factored form is given by the identity: - \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)\]. So, comparing \[{{x}^{2}}\] with a and \[{{y}^{2}}\] with b, we have,
\[\Rightarrow {{x}^{2}}{{y}^{3}}+{{y}^{9}}={{y}^{2}}\left[ {{x}^{2}}+{{y}^{2}} \right]\left[ {{x}^{4}}+{{y}^{4}}-{{x}^{2}}{{y}^{2}} \right]\]
Now, \[{{x}^{4}}\] can be written as \[{{\left( {{x}^{2}} \right)}^{2}}\] and similarly \[{{y}^{4}}\] can be written as \[{{\left( {{y}^{2}} \right)}^{2}}\], so we get,
\[\Rightarrow {{x}^{2}}{{y}^{3}}+{{y}^{9}}={{y}^{3}}\left[ {{x}^{2}}+{{y}^{2}} \right]\left[ {{\left( {{x}^{2}} \right)}^{2}}+{{\left( {{y}^{2}} \right)}^{2}}-{{x}^{2}}{{y}^{2}} \right]\]
Using the algebraic identity: - \[{{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab\], we get,
\[\begin{align}
  & \Rightarrow {{x}^{2}}{{y}^{3}}+{{y}^{9}}={{y}^{3}}\left[ {{x}^{2}}+{{y}^{2}} \right]\left[ {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-2{{x}^{2}}{{y}^{2}}-{{x}^{2}}{{y}^{2}} \right] \\
 & \Rightarrow {{x}^{2}}{{y}^{3}}+{{y}^{9}}={{y}^{3}}\left[ {{x}^{2}}+{{y}^{2}} \right]\left[ {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-3{{x}^{2}}{{y}^{2}} \right] \\
\end{align}\]
Here, \[3{{x}^{2}}{{y}^{2}}\] can be written as \[{{\left( \sqrt{3}xy \right)}^{2}}\] as a whole square, so we get,
\[\Rightarrow {{x}^{2}}{{y}^{3}}+{{y}^{9}}={{y}^{3}}\left[ {{x}^{2}}+{{y}^{2}} \right]\left[ {{\left( {{x}^{2}}+{{y}^{2}} \right)}^{2}}-{{\left( \sqrt{3}xy \right)}^{2}} \right]\]
Using the algebraic identity \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\], we have,
\[\Rightarrow {{x}^{2}}{{y}^{3}}+{{y}^{9}}={{y}^{3}}\left[ {{x}^{2}}+{{y}^{2}} \right]\left[ {{x}^{2}}+{{y}^{2}}-\sqrt{3}xy \right]\left[ {{x}^{2}}+{{y}^{2}}+\sqrt{3}xy \right]\]
Hence, the above relation in the R.H.S. is the factored form of the expression \[{{x}^{6}}{{y}^{3}}+{{y}^{9}}\].

Note: One may note that if you want you may simplify the term \[{{x}^{2}}+{{y}^{2}}\] further by using the algebraic identity: - \[{{a}^{2}}+{{b}^{2}}={{\left( a-b \right)}^{2}}+2ab\] or \[{{\left( a+b \right)}^{2}}-2ab\]. Actually, in this type of question we generally simplify or factor the expression up to an extent so that it cannot be simplified further. Note that we can simplify the expression \[{{x}^{2}}+{{y}^{2}}\] in complex factors also which is given as: - \[{{x}^{2}}+{{y}^{2}}=\left( x+iy \right)\left( x-iy \right)\] where \[{{i}^{2}}=-1\]. Here, no information is provided to us regarding real or imaginary factors.