how to find the particular solution for c, e, f?

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What is the general form of the particular solution guaranteed to exist of the linear nonhomogeneous recurrence relation

$a_n$=$6a_{n-1}$-$12a_{n-2}$+$8a_{n-3}$+F(n) if

- F(n)=$n^2$
- F(n)=$2^n$
- F(n)=$n2^n$
- F(n)=$(-2)^n$
- F(n)=$n^22^n$
- F(n)=$n^3(-2)^n$
- F(n)=3

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First solve the equation $\alpha^{3} - 6\alpha ^{2} +12\alpha -8=0$

So, here $\alpha=2,2,2$

Now, for particular solutions,

For $c)$ root $2$ has multiplicity $m=3$

So,

$a_{n}^{(p)} = n^{3}(P_{1}n + P_{0})2^{n}$

Put it in recurrence and find $P_1$ and $P_0$

For $e)$

$a_{n}^{(p)} = n^{3}(P_2n^2+P_{1}n + P_{0})2^{n}$

Put it in recurrence and find $P_2,P_1$ and $P_0$

For $f)$

$a_{n}^{(p)} = (P_3n^3+P_2n^2+P_{1}n + P_{0})(-2)^{n}$

Put it in recurrence and find $P_3,P_2,P_1$ and $P_0$

So, here $\alpha=2,2,2$

Now, for particular solutions,

For $c)$ root $2$ has multiplicity $m=3$

So,

$a_{n}^{(p)} = n^{3}(P_{1}n + P_{0})2^{n}$

Put it in recurrence and find $P_1$ and $P_0$

For $e)$

$a_{n}^{(p)} = n^{3}(P_2n^2+P_{1}n + P_{0})2^{n}$

Put it in recurrence and find $P_2,P_1$ and $P_0$

For $f)$

$a_{n}^{(p)} = (P_3n^3+P_2n^2+P_{1}n + P_{0})(-2)^{n}$

Put it in recurrence and find $P_3,P_2,P_1$ and $P_0$