Question

Let $D_1 = \begin{vmatrix} b+c & a & b \\ c+a & c & a \\ a+b & b & c \end{vmatrix}$ and

$D_2 = \begin{vmatrix} (b+c)^2 & a^2 & bc \\ (c+a)^2 & b^2 & ca \\ (a+b)^2 & c^2 & ab \end{vmatrix}$

The divisor which is common to both $D_1$ and $D_2$ is

• A. (a−b)
• B. (ab + bc + ca)
• C. a+b+c
• D. (c-a)

Answer 1

Answer: (C), (D)

a+b+c, c-a

Answer 2

Solution:

We want to show that both

$$D_1=\begin{vmatrix} b+c & a & b \\ c+a & c & a \\ a+b & b & c \end{vmatrix}\quad\text{and}\quad D_2=\begin{vmatrix} (b+c)^2 & a^2 & bc \\ (c+a)^2 & b^2 & ca \\ (a+b)^2 & c^2 & ab \end{vmatrix}$$

are divisible by a factor that vanishes when either

$$a+b+c=0\quad\text{or}\quad c-a=0.$$

Checking $a+b+c$:

Choose any numbers such that

$$a+b+c=0.$$

For example, take $a=1,\;b=2,\;c=-3$. Then every row of each determinant is “balanced” so that direct computation shows

$$D_1=0\quad\text{and}\quad D_2=0.$$

Thus, the linear factor $a+b+c$ divides both $D_1$ and $D_2$.

Checking $c-a$:

Next, set

$$c=a.$$

For any values of $a$ and $b$, substitution $c=a$ causes two rows to become identical in each determinant (or forces their dependence), and hence

$$D_1=0\quad\text{and}\quad D_2=0.$$

Thus, $(c-a)$ is also a factor of both determinants.

Testing the other options:

A similar test (for example, setting $a=b$ with $a=1,\;b=1,\;c\neq1$) shows that neither $(a-b)$ nor $(ab+bc+ca)$ always cause $D_1$ and $D_2$ to vanish.

Conclusion:

Both $\boxed{a+b+c}$ and $\boxed{(c-a)}$ are common divisors of $D_1$ and $D_2$.