a)\(n^3-7n-6\)
\(=n^3-3n^2+3n^2-9n+2n-6\)
\(=n^2\left(n-3\right)+3n\left(n-3\right)+2\left(n-3\right)\)
\(=\left(n^2+3n+2\right)\left(n-3\right)\)
\(=\left(n^2+2n+n+2\right)\left(n-3\right)\)
\(=\left[n\left(n+2\right)+\left(n+2\right)\right]\left(n-3\right)\)
\(=\left(n+1\right)\left(n+2\right)\left(n-3\right)\)
b)Vì \(A=0\) suy ra \(\left(n+1\right)\left(n+2\right)\left(n-3\right)=0\)
\(\Rightarrow\left[\begin{matrix}n+1=0\\n+2=0\\n-3=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}n=-1\\n=-2\\n=3\end{matrix}\right.\)
a. Ta có: a \(=\) \(n^3-7n-6\)
\(=n^3-n-6n-6\)
\(=n\times\left(n^2-1\right)-6\times\left(n-1\right)\)
\(=n\times\left(n-1\right)\times\left(n+1\right)-6\times\left(n-1\right)\)
\(=\left(n-1\right)\times\left(n^2+n-6\right)\)
\(=\left(n-1\right)\times\left(n^2+3n-2n-6\right)\)
\(=\left(n-1\right)\times\left(n\times\left(n-2\right)+3\times\left(n-2\right)\right)\)
\(=\left(n-1\right)\times\left(n-2\right)\times\left(n+3\right)\)
b. Ta có: \(n^3-7n-6=0\)
\(\Rightarrow\left(n-1\right)\times\left(n-2\right)\times\left(n+3\right)=0\)
\(\Rightarrow\left[\begin{matrix}n-1=0\\n-2=0\\n+3=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}n=1\\n=2\\n=-3\end{matrix}\right.\)
Vậy: Để a = 0 thì n \(\in\left\{-3;1;2\right\}\).
