How to solve the determinant without expanding? Here is the the determinant equation:
$$
  \begin{vmatrix}
   1 &amp; bc &amp; bc(b+c)\\
   1 &amp; ca &amp; ca(c+a)\\ 
   1 &amp; ab &amp; ab(a+b)\\
  \end{vmatrix} = abc\begin{vmatrix}
                      a &amp; 1 &amp; b+c\\
                      b &amp; 1 &amp; c+a\\
                      c &amp; 1 &amp; a+b\\
                     \end{vmatrix}=0
$$
I am new and beginner in matrices and determinants.So please explain the steps of  the answers.
It's a famous rule of determinants that


  If we multiply a row (column) of a matrix A by a number, the determinant of A will be multiplied by the same number.


Suppose we are comfortably in $\mathbb{R}$ and suppose also that $a,b,c \ne 0$, then multiply and divide the first row for $a$, the second for $b$ and the third for $c$
$$
  \frac{abc}{abc}\begin{vmatrix}
   1 &amp; bc &amp; bc(b+c)\\
   1 &amp; ca &amp; ca(c+a)\\ 
   1 &amp; ab &amp; ab(a+b)\\
  \end{vmatrix} = \frac{1}{abc}\begin{vmatrix}
                      a &amp; abc &amp; abc(b+c)\\
                      b &amp; abc &amp; abc(c+a)\\
                      c &amp; abc &amp; abc(a+b)\\
                     \end{vmatrix}
$$

Now you can conclude beacuse taking out the $(abc)^2$ factor that multiplies the  second and third column you get

$$
   abc\begin{vmatrix}
                      a &amp; 1&amp; (b+c)\\
                      b &amp; 1 &amp; (c+a)\\
                      c &amp; 1 &amp; (a+b)\\
                     \end{vmatrix}
$$

EDIT

At this point, to conclude that the determinant is $0$ you can simply use the Sarrus' rule or, without expanding, you can subtract the first row to the second and third row to get

$$
abc
\begin{vmatrix}
  a &amp; 1 &amp; b+c\\
  b-a &amp; 0 &amp; a-b\\
  c-a &amp; 0 &amp; a-c\\
\end{vmatrix} = abc[(a-b)(c-a) - (a-c)(b-a)] = 0
$$

Notice that in the header of the proof I supposed that $a,b,c \ne 0$, but now we can conclude even if one between $a,b,c = 0$, because the property is satisfied as well.
We have
$$\begin{vmatrix}
                      a &amp; 1 &amp; b+c\\
                      b &amp; 1 &amp; c+a\\
                      c &amp; 1 &amp; a+b\\
                     \end{vmatrix} = \begin{vmatrix}
                      a &amp; 1 &amp; a+b+c\\
                      b &amp; 1 &amp; b+c+a\\
                      c &amp; 1 &amp; c+a+b\\
                     \end{vmatrix} = 0$$
Here the first columm was added to the third. Thus it remains to verify the first equation in your question.

Case 1: $abc = 0$.

In that case at least one of $a, b ,c$ is $0$ and $$
 \begin{vmatrix}
   1 &amp; bc &amp; bc(b+c)\\
   1 &amp; ca &amp; ca(c+a)\\ 
   1 &amp; ab &amp; ab(a+b)\\
  \end{vmatrix}$$
has at least two rows of the form $(1 \quad 0 \quad 0)$ so that the first equation follows.

Case 2: $abc \ne 0$.

Then 
$$
 abc \begin{vmatrix}
   1 &amp; bc &amp; bc(b+c)\\
   1 &amp; ca &amp; ca(c+a)\\ 
   1 &amp; ab &amp; ab(a+b)\\
  \end{vmatrix} = \begin{vmatrix}
                      a &amp; abc &amp; abc(b+c)\\
                      b &amp; abc &amp; abc(c+a)\\
                      c &amp; abc &amp; abc(a+b)\\
                     \end{vmatrix} = (abc)^2\begin{vmatrix}
                      a &amp; 1 &amp; b+c\\
                      b &amp; 1 &amp; c+a\\
                      c &amp; 1 &amp; a+b\\
                     \end{vmatrix} 
$$
In the first step the factor $abc$ was shifted into the determinant by multiplying the first row with $a$, the second with $b$ and the third with $c$. In the second step the factor (abc) was extracted from columns two and three. 

Dividing by $abc \ne 0$ yields the desired result.