Question

Show that if $\frac{a}{b}=\frac{c}{d}=\frac{e}{f}$then show that each ratio is $\sqrt[3]{\frac{5a^3-2c^3+4e^3}{5b^3-2d^3+4f^3}}$.

Answer 1

$$\begin{gathered} \mathrm{Let}: \\ \frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k \\ \therefore a=kb,c=dkande=fk \\ Now,c\mathrm{onsider}\sqrt[3]{\frac{5a^3-2c^3+4e^3}{5b^3-2d^3+4f^3}}. \\ Putting\mathrm{the}\mathrm{values}\mathrm{of}a,c\mathrm{and}e,weget: \end{gathered}$$

$$\begin{gathered} =\sqrt[3]{\frac{5k^3{b}^3-2d^3{k}^3+4f^3{k}^3}{5b^3-2d^3+4f^3}} \\ =\sqrt[3]{k^3\times \frac{5b^3-2d^3+4f^3}{5b^3-2d^3+4f^3}} \\ =\sqrt[3]{k^3} \\ =k \\ \mathrm{Hence},\mathrm{each}\mathrm{ratio}\mathrm{is}\mathrm{equal}\mathrm{to}\sqrt[3]{\frac{5a^3-2c^3+4e^3}{5b^3-2d^3+4f^3}}. \end{gathered}$$