Question

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

Answer 1

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

$\left|\begin{array}{ccc}x+a& b& c\\ a& x+b& c\\ a& b& x+c\end{array}\right|=(x+a)\left[\right(x+b\left)\right(x+c)-bc]-b\left[a\right(x+c)-ac]+c[ab-a(x+b\left)\right]$

$=(x+a)(x^2+bx+cx+bc-bc)-b(ax+ac-ac)+c(ab-ax-ab)$

$=(x+a)(x^2+bx+cx)-bax-cax$

$=x^3+b{x}^2+c{x}^2+a{x}^2+abx+acx-bax-cax$

$=x^3+b{x}^2+c{x}^2+a{x}^2$

$=x^2(x+a+b+c)$

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

Hence, proved.