Show that $\left| {\begin{array}{*{20}{c}}
  1&{yz}&{y + z} \\
  1&{zx}&{z + x} \\
  1&{xy}&{x + y}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&x&{{x^2}} \\
  1&y&{{y^2}} \\
  1&z&{{z^2}}
\end{array}} \right|$.

Hint: First take the L.H.S. and subtract row 1 (${R_1}$) from row 2 (${R_2}$) and row 3 (${R_3}$). After that, take common from row 2 (${R_2}$) and row 3 (${R_3}$). Now, solve the determinant along column 1 (${C_1}$). Then, take the R.H.S. and subtract r and subtract row 1 (${R_1}$) from row 2 (${R_2}$) and row 3 (${R_3}$). After that, take common from row 2 (${R_2}$) and row 3 (${R_3}$). Now, solve the determinant along column 1 (${C_1}$). Now check if the value of L.H.S. is equal to R.H.S. or not.

Complete step-by-step answer:
Take L.H.S. and simplify it.
$\left| {\begin{array}{*{20}{c}}
  1&{yz}&{y + z} \\
  1&{zx}&{z + x} \\
  1&{xy}&{x + y}
\end{array}} \right|$
Subtract row 1 from row 2 and row 3,
$\begin{gathered}
  {R_2} \to {R_2} - {R_1} \\
  {R_3} \to {R_3} - {R_1} \\
\end{gathered} $
$\left| {\begin{array}{*{20}{c}}
  1&{yz}&{y + z} \\
  0&{zx - yz}&{\left( {z + x} \right) - \left( {y + z} \right)} \\
  0&{xy - yz}&{\left( {x + y} \right) - \left( {y + z} \right)}
\end{array}} \right|$
Take $z$ common from 2nd column of 2nd row, $y$ from 2nd column of 3rd row and simplify the 3rd column,
$\left| {\begin{array}{*{20}{c}}
  1&{yz}&{y + z} \\
  0&{z\left( {x - y} \right)}&{x - y} \\
  0&{y\left( {x - z} \right)}&{x - z}
\end{array}} \right|$
Take out $\left( {x - y} \right)$ common from 2nd row and $\left( {x - z} \right)$ from 3rd row.
$\left( {x - y} \right)\left( {x - z} \right)\left| {\begin{array}{*{20}{c}}
  1&{yz}&{y + z} \\
  0&z&1 \\
  0&y&1
\end{array}} \right|$
Solve the determinant along the first column (${C_1}$),
$\left( {x - y} \right)\left( {x - z} \right)\left[ {1\left( {\left| {\begin{array}{*{20}{c}}
  z&1 \\
  y&1
\end{array}} \right|} \right) - 0 + 0} \right]$
Solve the determinant inside the bracket,
$\left( {x - y} \right)\left( {x - z} \right)\left( {z - y} \right)$ ….. (1)
Now, take L.H.S. and simplify it.
$\left| {\begin{array}{*{20}{c}}
  1&x&{{x^2}} \\
  1&y&{{y^2}} \\
  1&z&{{z^2}}
\end{array}} \right|$
Subtract row 1 from row 2 and row 3,
$\begin{gathered}
  {R_2} \to {R_2} - {R_1} \\
  {R_3} \to {R_3} - {R_1} \\
\end{gathered} $
$\left| {\begin{array}{*{20}{c}}
  1&x&{{x^2}} \\
  0&{y - x}&{{y^2} - {x^2}} \\
  0&{z - x}&{{z^2} - {x^2}}
\end{array}} \right|$
Take out $\left( {y - x} \right)$ common from 2nd row and $\left( {z - x} \right)$ from 3rd row.
$\left( {y - x} \right)\left( {z - x} \right)\left| {\begin{array}{*{20}{c}}
  1&x&{{x^2}} \\
  0&1&{y + x} \\
  0&1&{z + x}
\end{array}} \right|$
Solve the determinant along the first column (${C_1}$),
$\left( {y - x} \right)\left( {z - x} \right)\left[ {1\left( {\left| {\begin{array}{*{20}{c}}
  1&{y + x} \\
  1&{z + x}
\end{array}} \right|} \right) - 0 + 0} \right]$
Solve the determinant inside the bracket,
$\left( {x - y} \right)\left( {x - z} \right)\left[ {\left( {z + x} \right) - \left( {y + x} \right)} \right]$
Solve the variables inside the bracket,
$\left( {x - y} \right)\left( {x - z} \right)\left( {z - y} \right)$ ….. (2)
Since the value of equation (1) and (2) are equal.
Hence, $\left| {\begin{array}{*{20}{c}}
  1&{yz}&{y + z} \\
  1&{zx}&{z + x} \\
  1&{xy}&{x + y}
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
  1&x&{{x^2}} \\
  1&y&{{y^2}} \\
  1&z&{{z^2}}
\end{array}} \right|$.

Note: Determinant, in linear and multilinear algebra, a value, denoted det A, associated with a square matrix A of n rows and n columns.
The determinant of a matrix A is commonly denoted det(A) or |A|
A $k \times k$ determinant $\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}& \cdots &{{a_{1k}}} \\
  {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2k}}} \\
   \vdots & \vdots & \ddots & \vdots \\
  {{a_{k1}}}&{{a_{k2}}}& \cdots &{{a_{kk}}}
\end{array}} \right|$ can be expanded "by minors" to obtain
\[{a_{11}}\left| {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}}& \cdots &{{a_{2k}}} \\
   \vdots & \vdots & \ddots & \vdots \\
  {{a_{k2}}}&{{a_{k3}}}& \cdots &{{a_{kk}}}
\end{array}} \right| - {a_{12}}\left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{23}}}& \cdots &{{a_{2k}}} \\
   \vdots & \vdots & \ddots & \vdots \\
  {{a_{k1}}}&{{a_{k3}}}& \cdots &{{a_{kk}}}
\end{array}} \right| + \ldots \pm {a_{1k}}\left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{22}}}& \cdots &{{a_{2\left( {k - 1} \right)}}} \\
   \vdots & \vdots & \ddots & \vdots \\
  {{a_{k1}}}&{{a_{k2}}}& \cdots &{{a_{k\left( {k - 1} \right)}}}
\end{array}} \right|\]