Suppose $a,b,c \neq 0$. Show that $\det(1+a,1,1),(1,1+b,1),(1,1,1+c) = abc\left(1 + \frac1a + \frac1b + \frac1c\right)$

Question

Suppose $a,b,c \neq 0$ .Show that the \begin{eqnarray*} \det \begin{bmatrix} 1+a &1 &1 \\1 &1+b & 1 \\ 1 & 1 & 1+c \end{bmatrix} = abc \left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right). \end{eqnarray*}

I've use cofactor expansion and reduced to get the form $bc + ac + ab + abc = det$. Choosing a variety of number to plug in for $a,b,$ and $c$, I know I'm on the right track. How can I reduce to get to $$bc + ac + ab + abc = abc\left(1 + \frac1a + \frac1b + \frac1c\right)$$

Answer 1

$$abc(1 + {1\over a} + {1\over b} + {1\over c})= abc \cdot 1 + abc \cdot {1\over a}+ abc \cdot {1\over b}+ abc \cdot {1\over c}= abc + bc + ac + ab$$