By using properties of determinants, show that:
$\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|=\left(1-x^{3}\right)^{2}$
$\Delta=\left|\begin{array}{ccc}1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3},$ we have:
$\Delta=\left|\begin{array}{ccc}1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1\end{array}\right|$
Applying $C_{2} \rightarrow C_{2}-C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1},$ we have:
$\Delta=\left(1+x+x^{2}\right)\left|\begin{array}{ccc}1 & 0 & 0 \\ x^{2} & 1-x^{2} & x-x^{2} \\ x & x^{2}-x & 1-x\end{array}\right|$
$=\left(1+x+x^{2}\right)(1-x)(1-x)\left|\begin{array}{ccc}1 & 0 & 0 \\ x^{2} & 1+x & x \\ x & -x & 1\end{array}\right|$
$=\left(1-x^{3}\right)(1-x)\left|\begin{array}{ccc}1 & 0 & 0 \\ x^{2} & 1+x & x \\ x & -x & 1\end{array}\right|$
Expanding along $R_{1},$ we have:
$\Delta=\left.\left(1-x^{3}\right)(1-x)(1)\right|_{-x} ^{1+x} \quad 1 |$
$=\left(1-x^{3}\right)(1-x)\left(1+x+x^{2}\right)$
$=\left(1-x^{3}\right)\left(1-x^{3}\right)$
$=\left(1-x^{3}\right)^{2}$
Hence, the given result is proved.
If $f(x) = \left| {\begin{array}{*{20}{c}}{x - 3}&{2{x^2} - 18}&{3{x^3} - 81}\\{x - 5}&{2{x^2} - 50}&{4{x^3} - 500}\\1&2&3\end{array}} \right|$ then $f(1).f(3) + f(3).f(5) + f(5).f(1)$=
The value of $\left| {\,\begin{array}{*{20}{c}}{{5^2}}&{{5^3}}&{{5^4}}\\{{5^3}}&{{5^4}}&{{5^5}}\\{{5^4}}&{{5^5}}&{{5^7}}\end{array}\,} \right|$ is
$2\,\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\a&b&c\\{{a^2} - bc}&{{b^2} - ac}&{{c^2} - ab}\end{array}\,} \right| = $
If ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r - 1}}}&{{{2.3}^{r - 1}}}&{{{4.5}^{r - 1}}}\\x&y&z\\{{2^n} - 1}&{{3^n} - 1}&{{5^n} - 1}\end{array}} \right|$, then the value of $\sum\limits_{r = 1}^n {{D_r} = } $
$\left| {\,\begin{array}{*{20}{c}}{{b^2} + {c^2}}&{{a^2}}&{{a^2}}\\{{b^2}}&{{c^2} + {a^2}}&{{b^2}}\\{{c^2}}&{{c^2}}&{{a^2} + {b^2}}\end{array}\,} \right| = $