Question

If $abc\ne 0$, then
$\Delta \left(x\right)=\left|\begin{array}{ccc}{a}^2(1+x)& ab& ac\\ ab& b^2(1+x)& bc\\ ac& bc& c^2(1+x)\end{array}\right|$
is divisible by

• A. $3+x$
• B. $x$
• C. $x^2$
• D. ${(1+x)}^2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $3+x$
B $x$
C $x^2$

$\Delta \left(x\right)=\left|\begin{array}{ccc}{a}^2(1+x)& ab& ac\\ ab& b^2(1+x)& bc\\ ac& bc& c^2(1+x)\end{array}\right|$

Taking $a,b,c$ common from $C_1,C_2,C_3$, we get

$\Delta \left(x\right)=abc\left|\begin{array}{ccc}a(1+x)& a& a\\ b& b(1+x)& b\\ c& c& c(1+x)\end{array}\right|$

Now taking $a,b,c$ common from $R_1,R_2,R_3$

$\Delta \left(x\right)=a^2{b}^2{c}^2\left|\begin{array}{ccc}(1+x)& 1& 1\\ 1& (1+x)& 1\\ 1& 1& 1(1+x)\end{array}\right|$

$\Delta \left(x\right)=a^2{b}^2{c}^2\left[\right(1+x\left)\right[(1+x{)}^2-1]-1\left[\right(1+x)-1]+1[1-(1+x\left)\right]]$

$\Rightarrow \Delta \left(x\right)=3a^2{b}^2{c}^2{x}^2(3+x)$