If \(\frac{x}{y}=\frac{z}{w}=c\). find the value of \(\frac{3 x^{2}-x z+z}{3 y-y x+w}\) in terms of \(c^{2} \)
Explanation
\(\frac{x}{y}=\frac{z}{w}=c\) (given equation \()\)
\(\frac{3 x^{2}-x z+z^{2}}{3 y^{2}-y w+x^{2}}=\) ? (required)
from the given equation, we have that
\(\frac{x}{y}=c \Rightarrow \frac{x^{2}}{y^{2}}=c^{2} \ldots \ldots \ldots \ldots \ldots(1)\)
\(\frac{z}{w}=c \cdot \frac{z^{2}}{w^{2}}=c^{2}\)
\(\frac{x}{y}=\frac{z}{w}\) or \(\frac{z}{x}=\frac{w}{y}=c\)
\(\Rightarrow \frac{z^{2}}{x^{2}}=\frac{w^{2}}{y^{2}}=c^{2}\)
i.e. \(\frac{z^{2}}{x^{2}}=c^{2}\) and \(\frac{w^{2}}{y^{2}}=c^{2}\)
from
\begin{aligned}
&\frac{3 x^{2}-x z+z^{2}}{3 y^{2}-y w+w^{2}} \\
&=\frac{x^{2}\left(3-\frac{z}{x}+\frac{z^{2}}{x^{2}}\right)}{y^{2}\left(3-\frac{w}{y}+\frac{w^{2}}{z^{2}}\right)}=\frac{x^{2}}{y^{2}}\left(\frac{3-\frac{z}{x}+\frac{z^{2}}{x^{2}}}{3-\frac{w}{y}+\frac{w^{2}}{z^{2}}}\right)
\end{aligned}
but from above, \(\frac{x^{2}}{y^{2}}=c^{2}, \frac{z}{x}=c\) \(\frac{w}{y}=c .\) substituting these we have
\(c^{2}\left(\frac{3-c+c^{2}}{3-c+c^{2}}\right)=c^{2}\)