Ta có:
\(Q\left(1\right)=a+b+c+d\Rightarrow a+b+c⋮3\left(1\right)\)
\(Q\left(-1\right)=-a+b-c+d⋮3\left(2\right)\)
Cộng (1) với (2), ta có: \(2b+2d⋮3\)
Mà \(d⋮3\Rightarrow2d⋮3\)
\(\Rightarrow2b⋮3\Rightarrow b⋮3\)
\(Q\left(2\right)=8a+4b+2c+d⋮3\)
\(\Rightarrow8a+2c⋮3\)(vì \(4b+d⋮3\))
\(\Rightarrow6a+2a+2c⋮3\)
\(\Rightarrow6a+2\left(a+c\right)⋮3\)
Mà \(a+c⋮3\left(a+b+c⋮3,b⋮3\right)\)
\(\Rightarrow6a⋮3\)
\(\Rightarrow a⋮3\)
\(\Rightarrow c⋮3\)
\(d⋮3\left(gt\right)\)