<s>[INST] Value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$, $P(1)=10$, $P(2)=20$, $P(3)=30$ 
What will be the value of $P(12)+P(-8)$ if $P(x)=x^{4}+ax^{3}+bx^{2}+cx+d$
  provided that $P(1)=10$, $P(2)=20$, $P(3)=30$?

I put these values and got three simultaneous equations in $a, b, c, d$. What is the smarter way to approach these problems?
 [/INST] Other way doing this:
We try to find reals $e,f,g$ such that $P(12)+P(-8)=eP(1)+fP(2)+gP(3)$. So, if we try to equal "$x^k$ evaluated", we gain a system of equations:
$$
\left\{\begin{array}{ccc}
1^ke+2^kf+3^kg&=&12^k+(-8)^k
\end{array}\right.,\quad k=0,\cdots,4
$$
In particular,
$$
\left\{\begin{array}{ccc}
e+f+g&=&1+1\\
e+2f+3g&=&12+(-8)\\
e+2^2f+3^2g&=&12^2+(-8)^2
\end{array}\right.
$$
and we obtain $e=100,f=-198,g=100$. Verifying the others values for $k$:
$$
\left\{\begin{array}{ccc}
100+2^3(-198)+3^3\cdot100&=&12^3+(-8)^3\\
100+2^4\cdot(-198)+3^4\cdot100&=&12^4+(-8)^4 + 19800
\end{array}\right.
$$
So, $P(12)+P(-8)=100P(1)-198P(2)+100P(3)+1980=19840$
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