Ta có \(P\left(1\right)=a+b+c+d=100\) (1)
\(P\left(-1\right)=-a+b-c+d=50\) (2)
\(P\left(0\right)=d=1\)mà \(a+b+c+d=100\)nên \(a+b+c=99\)
\(P\left(2\right)=8a+4b+2c+d=120\)
Từ (1) và (2) ta có
\(\left(a+b+c+d\right)+\left(-a+b-c+d\right)=100+50\Rightarrow2b+2d=150\)
\(\Rightarrow2b+2=150\Rightarrow2b=148\Rightarrow b=74\)
Ta có \(8a+4b+2c+d=120\Rightarrow6a+2b+\left(a+b+c\right)+\left(a+b+c+d\right)=120\)
\(\Rightarrow6a+2b+99+100=120\Rightarrow6a+2b+199=120\Rightarrow6a+148+199=120\)
\(\Rightarrow6a=-277\Rightarrow a=\frac{-277}{6}\)
Vì \(a+b+c=99\)mà \(a=-\frac{277}{6};b=74\)nên \(c=\frac{377}{6}\)
Khi đó \(P\left(x\right)=-\frac{277}{6}x^3+74x^2+\frac{377}{6}x+1\)
Do đó \(P\left(3\right)=\frac{-277}{6}.3^3+74.3^2+\frac{377}{6}.3+1=-833+666+1=-166\)
Vậy P(3)=-166
