Originally Posted by
Shyam For the second,
$\displaystyle = \left| {\begin{array}{*{20}c}
1 & 2 & {2^2 } & {2^3 } \\
1 & 3 & {3^2 } & {3^3 } \\
1 & 5 & {5^2 } & {5^3 } \\
1 & 6 & {6^2 } & {6^3 } \\
\end{array} } \right| \hfill \\$
$\displaystyle {\text{Apply }}R_2 \to R_2 - R_1 ,{\text{ }}R_3 \to R_3 - R_1 ,{\text{ }}R_4 \to R_4 - R_1 \hfill \\$
$\displaystyle = \left| {\begin{array}{*{20}c}
1 & 2 & {2^2 } & {2^3 } \\
0 & {3 - 2} & {3^2 - 2^2 } & {3^3 - 2^3 } \\
0 & {5 - 2} & {5^2 - 2^2 } & {5^3 - 2^3 } \\
0 & {6 - 2} & {6^2 - 2^2 } & {6^3 - 2^3 } \\
\end{array} } \right| \hfill \\$
$\displaystyle = \left| {\begin{array}{*{20}c}
{3 - 2} & {3^2 - 2^2 } & {3^3 - 2^3 } \\
{5 - 2} & {5^2 - 2^2 } & {5^3 - 2^3 } \\
{6 - 2} & {6^2 - 2^2 } & {6^3 - 2^3 } \\
\end{array} } \right| \hfill \\$
$\displaystyle {\text{Take common }}\left( {3 - 2} \right){\text{ from }}R_1 {\text{, }}\left( {5 - 2} \right){\text{ from }}R_2 {\text{ and }}\left( {6 - 2} \right){\text{ from }}R_3 \hfill \\$
$\displaystyle = \left( {3 - 2} \right)\left( {5 - 2} \right)\left( {6 - 2} \right)\left| {\begin{array}{*{20}c}
1 & {3 + 2} & {3^2 + 3.2 + 2^2 } \\
1 & {5 + 2} & {5^2 + 5.2 + 2^2 } \\
1 & {6 + 2} & {6^2 + 6.2 + 2^2 } \\
\end{array} } \right| \hfill \\$
$\displaystyle \left[ {{\text{since, }}a^2 - b^2 = \left( {a - b} \right)\left( {a + b} \right){\text{ and }}a^3 - b^3 = \left( {a - b} \right)\left( {a^2 + ab + b^2 } \right)} \right] \hfill \\$
$\displaystyle = 1 \times 3 \times 4\left| {\begin{array}{*{20}c}
1 & 5 & {19} \\
1 & 7 & {39} \\
1 & 8 & {52} \\
\end{array} } \right| \hfill \\$
$\displaystyle {\text{Now, apply }}R_2 \to R_2 - R_1 ,{\text{ }}R_3 \to R_3 - R_1 , \hfill \\$
Now, finish up. its value is 72