Question

If ${\Delta }_1=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}bc& b+c& 1\\ ca& c+a& 1\\ ab& a+b& 1\end{array}\right|$ then $\frac{{\Delta }_1}{{\Delta }_2}=$

• A. $ab+bc+ca$
• B. $abc$
• C. $2(ab+bc+ca)$
• D. $(a+b+c{)}^2$

Answer 1

Answer: (A)

${\Delta }_1=\left|\begin{array}{ccc}1& a^2& a^3\\ 1& b^2& b^3\\ 1& c^2& c^3\end{array}\right|$ $R_2\to R_2-R_1,R_3\to R_3-R_1$
$=\left|\begin{array}{ccc}1& a^2& a^3\\ 0& b^2-a^2& b^3-a^3\\ 0& c^2-a^2& c^3-a^3\end{array}\right|$
$(b-a)(c-a)\left|\begin{array}{ccc}1& a^2& a^3\\ 0& b+a& b^2+a^2+ab\\ 0& c+a& c^2+a^2+ac\end{array}\right|$
$=(b-a)(c-a)[(b+a)\left(c^2+a^2+ac\right)$ $-(c+a)\left(b^2+a^2+ab\right)]$
$(b-a)(c-a)[b{c}^2+a^{2}b+abc+a{c}^2$
$+a^3+a^{2}c-b^{2}c-a^{2}c-abc-b^{2}a-a^3-a^{2}b]$
$=(b-a)(c-a)[(b-c)(ab+bc+ca)]$
$=-(a-b)(b-c)(c-a)(ab+bc+ca)$
Now, ${\Delta }_2=\left|\begin{array}{ccc}bc& b+c& 1\\ ca& c+a& 1\\ ab& a+b& 1\end{array}\right|$
$R_2\to R_2-R_1$ and $R_3\to R_3-R_1$
$=\left|\begin{array}{ccc}bc& b+c& 1\\ c(a-b)& a-b& 0\\ b(c-a)& a-c& 0\end{array}\right|=(a-b)(c-a)\left|\begin{array}{ccc}bc& b+c& 1\\ c& 1& 0\\ a& 1& 0\end{array}\right|$
$=(a-b)(c-a)[l(c-b)]$
$=(a-b)(c-b)(c-a)$
Now, $\frac{{\Delta }_1}{{\Delta }_2}=\frac{-(a-b)(b-c)(c-a)(ab+bc+ca)}{-(a-b)(b-c)(c-a)}$
$\frac{{\Delta }_1}{{\Delta }_2}=(ab+bc+ca)$