Question

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

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

Answer 1

Answer: (A), (C)

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

b"b'It has multiple answers

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

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

$=(abc{)}^2\left|\begin{array}{ccc}1+n& 1& 1\\ 1& 1+x& 1\\ 1& 1& 1+x\end{array}\right|$

$=\left(abc{)}^2|\right(1+x\left)|\right(1+x{)}^2-1]-(1+x-1)+(1-1-x)$

$=\left(abc{)}^2\right[(1+x{)}^3-(1+x)-x-x]$

$=\left(abc{)}^2\right[(1+x{)}^3-1-x-2x]$

$=\left(abc{)}^2\right[1+x^3+3x(1+x)-1-3x]$

$=\left(abc{)}^2\right[x^3+3x+3x^2-3x]$

$=\left(abc{)}^2\right[x^2(x+3)]$

Divisible by $x^{2}b(x+3)$