Answer
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Hint:
Make the LHS of the first 3 equations the same. Then find the individual values of x, y and z. Substitute the values in the 4th equation, simplify the equation and you will get the answer without variables x, y and z.
Complete step-by-step answer:
We are given 4 equations,
\[\begin{align}
& yz={{a}^{2}}-(1) \\
& zx={{b}^{2}}-(2) \\
& xy={{c}^{2}}-(3) \\
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{d}^{2}}-(4) \\
\end{align}\]
The following 4 expressions are algebraic expressions built up from integers, constants, variables and the algebraic expression (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number).
Multiply ‘x’ on the LHS and RHS of equation (1).
Multiply ‘y’ on the LHS and RHS of equation (2).
Multiply ‘z’ on the LHS and RHS of equation (3).
\[\therefore \]Equation (1) becomes, \[xyz=x{{a}^{2}}\].
Equation (2) becomes, \[xyz=y{{b}^{2}}\].
Equation (3) becomes, \[xyz=z{{c}^{2}}\].
The LHS of all the above expressions are the same.
\[\therefore x{{a}^{2}}=y{{b}^{2}}=z{{c}^{2}}-(4)\]
Now we can put\[{{a}^{2}}x={{b}^{2}}y={{c}^{2}}z=k\], where k is a constant.
\[\begin{align}
& \therefore {{a}^{2}}x=k,{{b}^{2}}y=k,{{c}^{2}}z=k \\
& \therefore x=\dfrac{k}{{{a}^{2}}} \\
\end{align}\]
Similarly, \[y=\dfrac{k}{{{b}^{2}}}\]and \[z=\dfrac{k}{{{c}^{2}}}-(5)\]
So we find the values of x, y and z. Now substitute these values in equation (4).
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{d}^{2}} \\
& {{\left( \dfrac{k}{{{a}^{2}}} \right)}^{2}}+{{\left( \dfrac{k}{{{b}^{2}}} \right)}^{2}}+{{\left( \dfrac{k}{{{c}^{2}}} \right)}^{2}}={{d}^{2}} \\
& \Rightarrow \dfrac{{{k}^{2}}}{{{a}^{4}}}+\dfrac{{{k}^{2}}}{{{b}^{4}}}+\dfrac{{{k}^{2}}}{{{c}^{4}}}={{d}^{2}} \\
\end{align}\]
Take \[{{k}^{2}}\]common from LHS and simplify the LHS.
\[\begin{align}
& {{k}^{2}}\left[ \dfrac{{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}+{{a}^{4}}{{b}^{4}}}{{{a}^{4}}{{b}^{4}}{{c}^{4}}} \right]={{d}^{2}} \\
& \therefore {{k}^{2}}\left[ {{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}} \right]={{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}} \\
& \therefore {{k}^{2}}=\dfrac{{{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}}}{{{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}} \\
\end{align}\]
Taking square root on both sides.
\[\begin{align}
& \sqrt{{{k}^{2}}}=\sqrt{\dfrac{{{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}}}{{{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}}} \\
& k=\dfrac{{{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}}} \\
\end{align}\]
We know, \[y=\dfrac{k}{{{b}^{2}}}\].
\[\therefore y=\dfrac{d{{a}^{2}}{{b}^{2}}{{c}^{2}}}{{{b}^{2}}\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}=\dfrac{d{{a}^{2}}{{c}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}\]
Similarly, \[z=\dfrac{k}{{{c}^{2}}}\].
\[\therefore z=\dfrac{d{{a}^{2}}{{b}^{2}}{{c}^{2}}}{{{c}^{2}}\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}=\dfrac{d{{a}^{2}}{{b}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}\]
\[yz={{a}^{2}}\], substitute the values of y and z on this equation (1).
\[\begin{align}
& \left( \dfrac{d{{a}^{2}}{{c}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}} \right)\left( \dfrac{d{{a}^{2}}{{b}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}} \right)={{a}^{2}} \\
& \Rightarrow \dfrac{d{{a}^{4}}{{b}^{2}}{{c}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}={{a}^{2}} \\
\end{align}\]
Cancel out \[{{a}^{2}}\]on LHS and RHS.
\[\begin{align}
& \dfrac{d{{a}^{4}}{{b}^{2}}{{c}^{2}}}{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}=1 \\
& \therefore d{{a}^{4}}{{b}^{2}}{{c}^{2}}={{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}} \\
\end{align}\]
Hence we eliminated x, y, z and got the following expressions.
Note:
After getting the value of k, formulate the values in equation (5) and find the values of y and z. By this we can eliminate the constant ‘k’ which we have assumed. Then finally substitute the values of y and z in equation (1). We will get the final expression eliminating x, y and z.
Make the LHS of the first 3 equations the same. Then find the individual values of x, y and z. Substitute the values in the 4th equation, simplify the equation and you will get the answer without variables x, y and z.
Complete step-by-step answer:
We are given 4 equations,
\[\begin{align}
& yz={{a}^{2}}-(1) \\
& zx={{b}^{2}}-(2) \\
& xy={{c}^{2}}-(3) \\
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{d}^{2}}-(4) \\
\end{align}\]
The following 4 expressions are algebraic expressions built up from integers, constants, variables and the algebraic expression (addition, subtraction, multiplication, division and exponentiation by an exponent that is a rational number).
Multiply ‘x’ on the LHS and RHS of equation (1).
Multiply ‘y’ on the LHS and RHS of equation (2).
Multiply ‘z’ on the LHS and RHS of equation (3).
\[\therefore \]Equation (1) becomes, \[xyz=x{{a}^{2}}\].
Equation (2) becomes, \[xyz=y{{b}^{2}}\].
Equation (3) becomes, \[xyz=z{{c}^{2}}\].
The LHS of all the above expressions are the same.
\[\therefore x{{a}^{2}}=y{{b}^{2}}=z{{c}^{2}}-(4)\]
Now we can put\[{{a}^{2}}x={{b}^{2}}y={{c}^{2}}z=k\], where k is a constant.
\[\begin{align}
& \therefore {{a}^{2}}x=k,{{b}^{2}}y=k,{{c}^{2}}z=k \\
& \therefore x=\dfrac{k}{{{a}^{2}}} \\
\end{align}\]
Similarly, \[y=\dfrac{k}{{{b}^{2}}}\]and \[z=\dfrac{k}{{{c}^{2}}}-(5)\]
So we find the values of x, y and z. Now substitute these values in equation (4).
\[\begin{align}
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{d}^{2}} \\
& {{\left( \dfrac{k}{{{a}^{2}}} \right)}^{2}}+{{\left( \dfrac{k}{{{b}^{2}}} \right)}^{2}}+{{\left( \dfrac{k}{{{c}^{2}}} \right)}^{2}}={{d}^{2}} \\
& \Rightarrow \dfrac{{{k}^{2}}}{{{a}^{4}}}+\dfrac{{{k}^{2}}}{{{b}^{4}}}+\dfrac{{{k}^{2}}}{{{c}^{4}}}={{d}^{2}} \\
\end{align}\]
Take \[{{k}^{2}}\]common from LHS and simplify the LHS.
\[\begin{align}
& {{k}^{2}}\left[ \dfrac{{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}+{{a}^{4}}{{b}^{4}}}{{{a}^{4}}{{b}^{4}}{{c}^{4}}} \right]={{d}^{2}} \\
& \therefore {{k}^{2}}\left[ {{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}} \right]={{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}} \\
& \therefore {{k}^{2}}=\dfrac{{{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}}}{{{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}} \\
\end{align}\]
Taking square root on both sides.
\[\begin{align}
& \sqrt{{{k}^{2}}}=\sqrt{\dfrac{{{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}}}{{{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}}} \\
& k=\dfrac{{{d}^{2}}{{a}^{4}}{{b}^{4}}{{c}^{4}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{b}^{4}}{{c}^{4}}+{{a}^{4}}{{c}^{4}}}} \\
\end{align}\]
We know, \[y=\dfrac{k}{{{b}^{2}}}\].
\[\therefore y=\dfrac{d{{a}^{2}}{{b}^{2}}{{c}^{2}}}{{{b}^{2}}\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}=\dfrac{d{{a}^{2}}{{c}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}\]
Similarly, \[z=\dfrac{k}{{{c}^{2}}}\].
\[\therefore z=\dfrac{d{{a}^{2}}{{b}^{2}}{{c}^{2}}}{{{c}^{2}}\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}=\dfrac{d{{a}^{2}}{{b}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}\]
\[yz={{a}^{2}}\], substitute the values of y and z on this equation (1).
\[\begin{align}
& \left( \dfrac{d{{a}^{2}}{{c}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}} \right)\left( \dfrac{d{{a}^{2}}{{b}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}} \right)={{a}^{2}} \\
& \Rightarrow \dfrac{d{{a}^{4}}{{b}^{2}}{{c}^{2}}}{\sqrt{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}}={{a}^{2}} \\
\end{align}\]
Cancel out \[{{a}^{2}}\]on LHS and RHS.
\[\begin{align}
& \dfrac{d{{a}^{4}}{{b}^{2}}{{c}^{2}}}{{{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}}}=1 \\
& \therefore d{{a}^{4}}{{b}^{2}}{{c}^{2}}={{a}^{4}}{{b}^{4}}+{{c}^{4}}{{a}^{4}}+{{b}^{4}}{{c}^{4}} \\
\end{align}\]
Hence we eliminated x, y, z and got the following expressions.
Note:
After getting the value of k, formulate the values in equation (5) and find the values of y and z. By this we can eliminate the constant ‘k’ which we have assumed. Then finally substitute the values of y and z in equation (1). We will get the final expression eliminating x, y and z.