<s>[INST] find $f(x)$ when $3f(x-6)-2f(x-9)=x^2-54$ I can easily show that with the assumption $f$ is a polynomial $f(x)=x^2$. But without that assumption how can I prove that $f(x)=x^2$???. I have tried many change of variables $x=u+k$ but to no result. I am lost here
 [/INST] Let $f(x)$ be polynomial of $x$.
Let $n\ge 3,$ be the smallest power of $x$ whose coefficient$(a)$ may be $\ne 0$.
The coefficient of $x^n$ in $3f(x−6)−2f(x−9)$ is $3a-2a=a$
But the coefficient of $x^n$ where $n\ge 3$ in $x^2−54$ is $0\implies a=0$
So, $f(x)$ can not contain any higher powers $(\ge 3)$ of $x.$
So, $f(x)=bx^2+cx+d$
$3f(x−6)−2f(x−9)=x^2−54$
$b(x-h)^2+c(x-h)+d=x^2(b)+x(c-2bh)+(d-ch+bh^2)$
The coefficient of $x^2$ in $3f(x−6)−2f(x−9)$ is $3b-2b=b$
But the coefficient of $x^2$ in $x^2−54$ is $1\implies b=1$
The coefficient of $x$ in $f(x−h)$ is $c-2bh=c-2h$
So, the coefficient of $x$ in $3f(x−6)−2f(x−9)$ is $3\{c-2(-6)\}-2\{c-2(-9)\}=c$
But the coefficient of $x$ in $x^2−54$ is $0\implies c=0$
The constant term  in $f(x−h)$ is $d-ch+bh^2=d+h^2$
The constant term  in $3f(x−6)−2f(x−9$ is $3\{d+(-6)^2\}-2\{d+(-9)^2\}=d-54$
The constant term  in $x^2−54$ is $-54\implies d-54=-54\implies d=0$
So, $f(x)=x^2+0x+0=x^2$
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