How do you factor $18{{x}^{6}}+5{{y}^{6}}$?

Hint: To factor the given polynomial, we have to write it as the sum of two cubes so that we can use the algebraic identity given by \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\] in order to factor it. For this, we have to write \[18{{x}^{6}}\] as \[{{\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}} \right)}^{3}}\] and $5{{y}^{6}}$ as ${{\left( {{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)}^{3}}$ in the given polynomial. After factoring the given polynomial, we will require the exponent property given by ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, in order to simplify it.

Complete step by step solution:
The given polynomial is in the terms of two independent variables, x and y. Therefore, we can write the polynomial given in the above question as
$\Rightarrow p\left( x,y \right)=18{{x}^{6}}+5{{y}^{6}}$
The above expression can be written as the sum of two cubes by writing \[18{{x}^{6}}\] as \[{{\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}} \right)}^{3}}\], and $5{{y}^{6}}$ as ${{\left( {{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)}^{3}}$ in the above expression to get
$\Rightarrow p\left( x,y \right)={{\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}} \right)}^{3}}+{{\left( {{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)}^{3}}$
Now, we can solve the sum of cubes by using the algebraic identity given by \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\]. So the above expression can be written as
\[\begin{align}
  & \Rightarrow p\left( x,y \right)=\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}}+{{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)\left[ {{\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}} \right)}^{2}}-\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}} \right)\left( {{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)+{{\left( {{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)}^{2}} \right] \\
 & \Rightarrow p\left( x,y \right)=\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}}+{{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)\left[ {{18}^{\dfrac{2}{3}}}{{x}^{4}}-\left( {{18}^{\dfrac{1}{3}}} \right)\left( {{5}^{\dfrac{1}{3}}} \right){{x}^{2}}{{y}^{2}}+{{5}^{\dfrac{2}{3}}}{{y}^{4}} \right] \\
\end{align}\]
Now, by using the property of the multiplication of same exponents, which is given by ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, we can write the above expression as
\[\begin{align}
  & \Rightarrow p\left( x,y \right)=\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}}+{{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)\left[ {{18}^{\dfrac{2}{3}}}{{x}^{4}}-{{\left( 18\times 5 \right)}^{\dfrac{1}{3}}}{{x}^{2}}{{y}^{2}}+{{5}^{\dfrac{2}{3}}}{{y}^{4}} \right] \\
 & \Rightarrow p\left( x,y \right)=\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}}+{{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)\left[ {{18}^{\dfrac{2}{3}}}{{x}^{4}}-{{90}^{\dfrac{1}{3}}}{{x}^{2}}{{y}^{2}}+{{5}^{\dfrac{2}{3}}}{{y}^{4}} \right] \\
\end{align}\]
Hence, we have factored the given polynomial as \[\left( {{18}^{\dfrac{1}{3}}}{{x}^{2}}+{{5}^{\dfrac{1}{3}}}{{y}^{2}} \right)\left[ {{18}^{\dfrac{2}{3}}}{{x}^{4}}-{{90}^{\dfrac{1}{3}}}{{x}^{2}}{{y}^{2}}+{{5}^{\dfrac{2}{3}}}{{y}^{4}} \right]\].

Note: We might argue that $5$ and $18$ are not perfect cubes, nor do they have any common factor, so the given expression cannot be factored. But the question has directed us to do the factoring of the given polynomial $18{{x}^{6}}+5{{y}^{6}}$. It did not impose a condition that the coefficients in the factors must be rational numbers. Also, we must remember the algebraic identity \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\] carefully and must not forget the negative sign in the second factor \[\left( {{a}^{2}}-ab+{{b}^{2}} \right)\].