$$\begin{vmatrix} a-x& b & c\\1& -x& 0\\0& 1& -x \\\end{vmatrix} $$
I did
$$(a-x) \begin{vmatrix} -x& 0\\1&-x\end{vmatrix}-b\begin{vmatrix} 1& 0\\0&-x\end{vmatrix}+c\begin{vmatrix} 1& -x\\0&1\end{vmatrix}$$
I got $$ax^2-a-x^3+x+bx+c+cx$$
But the answer in the book is $$-x^3+ax^2+bx+c$$
Is the answer simplified or am I doing it incorrectly?
On
First the solution in the book is correct. In the future you can use something similar to the following to check: Woflram.
It looks like you did something wrong with your expansion. Remember:
$$\det \begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc.$$
You should have: $$(a-x)(x^2) -b(-x) + c$$
On
You expanded by the first row to get
$$(a-x) \begin{vmatrix} -x& 0\\1&-x\end{vmatrix}-b\begin{vmatrix} 1& 0\\0&-x\end{vmatrix}+c\begin{vmatrix} 1& -x\\0&1\end{vmatrix}$$
Now the first $2\times 2$ determinant equals $x^2$, the second $-x$ and the third $1$ and your expansion equals
$$(a-x)x^2+bx+c=-x^3+ax^2+bx+c$$
as expected.
On
$$(a-x) \begin{vmatrix} -x& 0\\1&-x\end{vmatrix}-b\begin{vmatrix} 1& 0\\0&-x\end{vmatrix}+c\begin{vmatrix} 1& -x\\0&1\end{vmatrix}$$
$$=(a-x)\left[(-x)(-x)-1\cdot0\right]-b\left[1\cdot(-x)-0\cdot 0\right]+c\left[1 \cdot 1-0\cdot (-x) \right]$$
$$=(a-x)x^2+bx+c$$
Can you complete it from here?
On
Just add this to point that you can expand by any row or column youwant, if it makes life simpler. Here, you can expand by the first column: $$\begin{vmatrix}a-x &b&c\\1&-x&0\\0&1&-x\end{vmatrix}= (a-x)\begin{vmatrix}-x&0\\1&-x\end{vmatrix}-\begin{vmatrix}b&c\\1&-x\end{vmatrix}=(a-x)x^2-(-bx-c)=\dotsm$$
Not quite - you seem to have evaluated some of the determinants for the $2\times 2$ matrices incorrectly. Your method for expanding the $3\times 3$ matrices is fine. But note $$\begin{vmatrix} -x& 0\\1&-x\end{vmatrix}=(-x)(-x)-(0)(1)=x^2\\\begin{vmatrix} 1& -x\\0&1\end{vmatrix}=(1)(1)-(-x)(0)=1$$ This will get rid of the extra $-a$, $+x$ and $+cx$ terms.