How do you factor $13{{\left( {{x}^{6}}+1 \right)}^{4}}\left( 18{{x}^{5}} \right){{\left( 9x+2 \right)}^{3}}+9{{\left( 9x+2 \right)}^{2}}\left( 9 \right){{\left( {{x}^{6}}+1 \right)}^{5}}$?

Hint: We first try to take common polynomials out from the given equation $13{{\left( {{x}^{6}}+1 \right)}^{4}}\left( 18{{x}^{5}} \right){{\left( 9x+2 \right)}^{3}}+9{{\left( 9x+2 \right)}^{2}}\left( 9 \right){{\left( {{x}^{6}}+1 \right)}^{5}}$. We take the polynomials or the factors with their maximum possible indices’ values. We find the simplified forms and then take their addition. From the addition we find the factorisation for the equation.

Complete step-by-step solution:
We have to find the factorisation of $13{{\left( {{x}^{6}}+1 \right)}^{4}}\left( 18{{x}^{5}} \right){{\left( 9x+2 \right)}^{3}}+9{{\left( 9x+2 \right)}^{2}}\left( 9 \right){{\left( {{x}^{6}}+1 \right)}^{5}}$.
There are two terms in addition. Every particular term is in multiplied form of polynomials.
We have to find the common polynomials from $13{{\left( {{x}^{6}}+1 \right)}^{4}}\left( 18{{x}^{5}} \right){{\left( 9x+2 \right)}^{3}}+9{{\left( 9x+2 \right)}^{2}}\left( 9 \right){{\left( {{x}^{6}}+1 \right)}^{5}}$.
Those are $\left( {{x}^{6}}+1 \right);\left( 9x+2 \right)$ with maximum possible indices value that we can take as common is 4 and 2 respectively.
We can also take constants as common from the equation where the maximum possible number is 9.
So, the polynomial that we are taking as common is $9{{\left( {{x}^{6}}+1 \right)}^{4}}{{\left( 9x+2 \right)}^{2}}$.
We are taking $9{{\left( {{x}^{6}}+1 \right)}^{4}}{{\left( 9x+2 \right)}^{2}}$ as common from the numbers $13{{\left( {{x}^{6}}+1 \right)}^{4}}\left( 18{{x}^{5}} \right){{\left( 9x+2 \right)}^{3}}$ and $9{{\left( 9x+2 \right)}^{2}}\left( 9 \right){{\left( {{x}^{6}}+1 \right)}^{5}}$.
If we take $9{{\left( {{x}^{6}}+1 \right)}^{4}}{{\left( 9x+2 \right)}^{2}}$ from $13{{\left( {{x}^{6}}+1 \right)}^{4}}\left( 18{{x}^{5}} \right){{\left( 9x+2 \right)}^{3}}$, the remaining number will be $\dfrac{13{{\left( {{x}^{6}}+1 \right)}^{4}}\left( 18{{x}^{5}} \right){{\left( 9x+2 \right)}^{3}}}{9{{\left( {{x}^{6}}+1 \right)}^{4}}{{\left( 9x+2 \right)}^{2}}}=13\left( 2{{x}^{5}} \right)\left( 9x+2 \right)$.
If we take $9{{\left( {{x}^{6}}+1 \right)}^{4}}{{\left( 9x+2 \right)}^{2}}$ from $9{{\left( 9x+2 \right)}^{2}}\left( 9 \right){{\left( {{x}^{6}}+1 \right)}^{5}}$, the remaining number will be $\dfrac{9{{\left( 9x+2 \right)}^{2}}\left( 9 \right){{\left( {{x}^{6}}+1 \right)}^{5}}}{9{{\left( {{x}^{6}}+1 \right)}^{4}}{{\left( 9x+2 \right)}^{2}}}=9\left( {{x}^{6}}+1 \right)$.
We have to take the addition of the terms of these simplified forms.
$13\left( 2{{x}^{5}} \right)\left( 9x+2 \right)=234{{x}^{6}}+52{{x}^{5}}$ and $9\left( {{x}^{6}}+1 \right)=9{{x}^{6}}+9$.
The addition gives $234{{x}^{6}}+52{{x}^{5}}+9{{x}^{6}}+9=243{{x}^{6}}+52{{x}^{5}}+9$
Therefore, the factorisation is \[9{{\left( {{x}^{6}}+1 \right)}^{4}}{{\left( 9x+2 \right)}^{2}}\left( 243{{x}^{6}}+52{{x}^{5}}+9 \right)\].

Note: The common terms are itself polynomial in their forms. We have to find the highest value for the indices. For this we take the minimum value of the indices for individual terms and that can be regarded as the maximum possible number for factorisation.