Question

If $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$then show that each ratio is ${\left(\frac{7a^4-3c^4+5e^4}{7b^4-3d^4+5f^4}\right)}^{\frac{1}{4}}.$

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k \\ \therefore a=bk,c=dk\mathrm{and}e=fk \\ \mathrm{Now},\mathrm{consider}{\left(\frac{7a^4-3c^4+5e^4}{7b^4-3d^4+5f^4}\right)}^{\frac{1}{4}}. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ {\left(\frac{7b^4{k}^4-3d^4{k}^4+5f^4{k}^4}{7b^4-3d^4+5f^4}\right)}^{\frac{1}{4}} \\ ={\left(\frac{k^4\left[7b^4-3d^4+5f^4\right]}{7b^4-3d^4+5f^4}\right)}^{\frac{1}{4}} \\ =k \\ \mathrm{We}\mathrm{know}\mathrm{that}\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k. \\ \mathrm{Hence},\mathrm{each}\mathrm{ratio}\mathrm{is}{\left(\frac{7a^4-3c^4+5e^4}{7b^4-3d^4+5f^4}\right)}^{\frac{1}{4}}. \end{gathered}$$