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# Show that $\left| \begin{matrix} 1 & 1 & 1 \\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\ {{x}^{3}} & {{y}^{3}} & {{z}^{3}} \\\end{matrix} \right|=\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( xy+yz+zx \right)$.

Last updated date: 23rd Jul 2024
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Hint: We must perform the following two column operations, ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$ and ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$. Then, by using the expansion formulae ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$, we can simplify the determinant to prove that it is equal to the given expression.

Complete step-by-step solution:
Let us assume a variable D that is equal to the given determinant, that is,
$D=\left| \begin{matrix} 1 & 1 & 1 \\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\ {{x}^{3}} & {{y}^{3}} & {{z}^{3}} \\ \end{matrix} \right|$
We also know that we can perform any row or column operation, without changing the value of determinant.
So, let us perform the column operation${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$. We now get,
$D=\left| \begin{matrix} 1 & 0 & 1 \\ {{x}^{2}} & {{y}^{2}}-{{z}^{2}} & {{z}^{2}} \\ {{x}^{3}} & {{y}^{3}}-{{z}^{3}} & {{z}^{3}} \\ \end{matrix} \right|$
Now, let us perform another column operation ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$. Thus, we will get
$D=\left| \begin{matrix} 1 & 0 & 0 \\ {{x}^{2}} & {{y}^{2}}-{{z}^{2}} & {{z}^{2}}-{{x}^{2}} \\ {{x}^{3}} & {{y}^{3}}-{{z}^{3}} & {{z}^{3}}-{{x}^{3}} \\ \end{matrix} \right|$
We know the expansion formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Thus, we can easily write
${{y}^{2}}-{{z}^{2}}=\left( y+z \right)\left( y-z \right)$ and ${{z}^{2}}-{{x}^{2}}=\left( z+x \right)\left( z-x \right)$.
We are also well aware about the expansion formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$.
Hence, we can write ${{y}^{3}}-{{z}^{3}}=\left( y-z \right)\left( {{y}^{2}}+{{z}^{2}}+yz \right)$ and ${{z}^{3}}-{{x}^{3}}=\left( z-x \right)\left( {{z}^{2}}+{{x}^{2}}+xz \right)$.
We can now substitute the value of all these expansions into the determinant. Thus, we get
$D=\left| \begin{matrix} 1 & 0 & 0 \\ {{x}^{2}} & \left( y+z \right)\left( y-z \right) & \left( z+x \right)\left( z-x \right) \\ {{x}^{3}} & \left( y-z \right)\left( {{y}^{2}}+{{z}^{2}}+yz \right) & \left( z-x \right)\left( {{z}^{2}}+{{x}^{2}}+xz \right) \\ \end{matrix} \right|$
We know that we can factor out the common terms from all the elements of a row or all the elements of a column.
We can see that the term $\left( y-z \right)$ can be taken as common from column 2. So, we get
$D=\left( y-z \right)\left| \begin{matrix} 1 & 0 & 0 \\ {{x}^{2}} & \left( y+z \right) & \left( z+x \right)\left( z-x \right) \\ {{x}^{3}} & \left( {{y}^{2}}+{{z}^{2}}+yz \right) & \left( z-x \right)\left( {{z}^{2}}+{{x}^{2}}+xz \right) \\ \end{matrix} \right|$
Also, we can see that the term $\left( z-x \right)$ can be taken as common from column 3. Hence, we get
$D=\left( y-z \right)\left( z-x \right)\left| \begin{matrix} 1 & 0 & 0 \\ {{x}^{2}} & \left( y+z \right) & \left( z+x \right) \\ {{x}^{3}} & \left( {{y}^{2}}+{{z}^{2}}+yz \right) & \left( {{z}^{2}}+{{x}^{2}}+xz \right) \\ \end{matrix} \right|$
Let us now solve the determinant. We can write
$D=\left( y-z \right)\left( z-x \right)\left[ 1\left\{ \left( y+z \right)\left( {{z}^{2}}+{{x}^{2}}+xz \right)-\left( z+x \right)\left( {{y}^{2}}+{{z}^{2}}+yz \right) \right\}-0+0 \right]$
Thus, we get
$D=\left( y-z \right)\left( z-x \right)\left[ \left( y+z \right)\left( {{z}^{2}}+{{x}^{2}}+xz \right)-\left( z+x \right)\left( {{y}^{2}}+{{z}^{2}}+yz \right) \right]$
We can multiply the expressions to get
$D=\left( y-z \right)\left( z-x \right)\left[ y{{z}^{2}}+{{x}^{2}}y+xyz+{{z}^{3}}+{{x}^{2}}z+x{{z}^{2}}-{{y}^{2}}z-{{z}^{3}}-y{{z}^{2}}-x{{y}^{2}}-x{{z}^{2}}-xyz \right]$
On simplification and cancelling the terms, we get
$D=\left( y-z \right)\left( z-x \right)\left[ {{x}^{2}}y+{{x}^{2}}z-{{y}^{2}}z-x{{y}^{2}} \right]$
We can rearrange these terms as follows,
$D=\left( y-z \right)\left( z-x \right)\left[ {{x}^{2}}y-x{{y}^{2}}+{{x}^{2}}z-{{y}^{2}}z \right]$
Thus, we can write
$D=\left( y-z \right)\left( z-x \right)\left[ xy\left( x-y \right)+z\left( {{x}^{2}}-{{y}^{2}} \right) \right]$
Again, using the expansion formula, we get
$D=\left( y-z \right)\left( z-x \right)\left[ xy\left( x-y \right)+z\left( x+y \right)\left( x-y \right) \right]$
Hence, we get
$D=\left( x-y \right)\left( y-z \right)\left( z-x \right)\left[ xy+z\left( x+y \right) \right]$
Or, we can write this as
$D=\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( xy+yz+zx \right)$
Hence, we have proved that $\left| \begin{matrix} 1 & 1 & 1 \\ {{x}^{2}} & {{y}^{2}} & {{z}^{2}} \\ {{x}^{3}} & {{y}^{3}} & {{z}^{3}} \\ \end{matrix} \right|=\left( x-y \right)\left( y-z \right)\left( z-x \right)\left( xy+yz+zx \right)$.

Note: We must note the difference in the symbols of determinant and matrix. We also know that when we take something as common from a matrix, is taken from each and every element of that matrix. But, in case of determinant, the common is taken only from a row or column.