If $${ \left( 1-{ x }^{ 3 } \right) }^{ n }=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r }{ \left( 1-x \right) }^{ 3n-2r } } $$ then find $a_r$.
My first attempt:
I wrote the above equation as:
$$\sum _{ i=0 }^{ n }{ { \left( -1 \right) }^{ n }{ x }^{ 3n }\left( \begin{matrix} n \\ i \end{matrix} \right) } =\sum _{ r=0 }^{ n }{ \sum _{ i=0 }^{ 3n-2r }{ { \left( -1 \right) }^{ i }{ a }_{ r }{ x }^{ r+i }\left( \begin{matrix} 3n-2r \\ i \end{matrix} \right) } } $$
From here, I couldn't proceed.
My second attempt:
I substituted $x=-1$ and did some algebraic bashing and got ${ a }_{ r }=\left( \begin{matrix} n \\ r \end{matrix} \right) { 3 }^{ r }$.
The answer I got was correct but I feel my method wasn't the expected one. Is there a method to find $a_r$ without any particular substitution?
Note $$(1-x^3)^n=(3x(1-x)+(1-x)^3)^n=\sum _{ r=0 }^{ n }{ 3^r\binom{n }{ r }{ [x(1-x) ]}^{ r }{ \left( 1-x \right) }^{ 3n-3r } }=\sum _{ r=0 }^{ n }{ 3^r\binom{n }{ r }{ x }^{ r }{ \left( 1-x \right) }^{ 3n-2r } } $$