Question

If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$ then show that $\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}$

Answer 1

$$\begin{gathered} Let: \\ \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k \\ \therefore a=bk,b=ckandc=dk \\ \Rightarrow a=d{k}^3,b=d{k}^{2}andc=dk \\ Now, \\ \mathrm{LHS}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{d^3{k}^9+d^3{k}^6+d^3{k}^3}{d^3{k}^6+d^3{k}^3+d^3}\left(S\mathrm{ubstituting}\mathrm{the}\mathrm{values}\mathrm{of}a,b\mathrm{and}c\right) \end{gathered}$$

$=\frac{d^3{k}^3\left(k^6+k^3+1\right)}{d^3\left(k^6+k^3+1\right)}$ = $k^3$

$$\begin{gathered} \mathrm{RHS}=\frac{a}{d} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}a=d{k}^3,\mathrm{we}\mathrm{get}: \\ \frac{d{k}^3}{d}=k^3 \end{gathered}$$

∴ LHS = RHS