Question

Evaluate: $\left|\begin{array}{cccc}1& a& a^2& a^3+bcd\\ 1& b& b^2& b^3+cda\\ 1& c& c^2& c^3+abd\\ 1& d& d^2& d^3+abc\end{array}\right|$

Answer 1

Split into two determinants
$\therefore △=$

$\left|\begin{array}{cccc}1& a& a^2& a^3\\ 1& b& b^2& b^3\\ 1& c& c^2& c^3\\ 1& d& d^2& d^3\end{array}\right|$ + $\left|\begin{array}{cccc}1& a& a^2& bcd\\ 1& b& b^2& cda\\ 1& c& c^2& dab\\ 1& d& d^2& abc\end{array}\right|$
= ${△}_1+{△}_2$
In ${△}_2$ multiply $R_1,R_2,R_3,R_4$ by a, b, c, d respectively and hence divide it by abcd
$\therefore {△}_2=\frac{1}{abcd}$$\left|\begin{array}{cccc}a& a^2& a^3& abcd\\ b& b^2& b^3& abcd\\ c& c^2& c^3& abcd\\ d& d^2& d^3& abcd\end{array}\right|$
$=\frac{abcd}{abcd}$$\left|\begin{array}{cccc}a& a^2& a^3& 1\\ b& b^2& b^3& 1\\ c& c^2& c^3& 1\\ d& d^2& d^3& 1\end{array}\right|$
Now cross the 4th column over three columns and hence write $(-1{)}^3$
$\therefore {△}_2=(-1{)}^3$$\left|\begin{array}{cccc}1& a& a^2& a^3\\ 1& b& b^2& c^2\\ 1& c& c^2& c^3\\ 1& d& d^2& d^3\end{array}\right|$ =-${△}_1$
$\therefore △={△}_2+{△}_1=0$