Question
If $\frac{a}{b}=\frac{c}{d}=\frac{r}{f}$, prove that $\left(\frac{a^2 b^2+c^2 d^2+e^2 f^2}{a b^3+c d^3+e f^3}\right)^{\frac{3}{2}}=\sqrt{\frac{a c e}{b d f}}$
✓
Answer
$\text { Let } \frac{a}{b}=\frac{c}{d}=\frac{r}{f}=k $
$\therefore a = bk , c = dk \text {, e }= fk $
$\text { L.H.S. } $
$=\left(\frac{a^2 b^2+c^2 d^2+e^2 f^2}{a b^3+c d^3+e f^3}\right)^{\frac{3}{2}} $
$=\left(\frac{b^2 k^2 \cdot b^2+d^2 k^2 \cdot d^2+f^2 k^2 \cdot f^2}{b k \cdot b^3+d k \cdot d^3+f k \cdot f^3}\right)^{\frac{3}{2}} $
$=\left[\frac{k^2\left(b^4+d^4+f^4\right)}{k\left(b^4+d^4+f^4\right)}\right]^{\frac{3}{2}} $
$=k^{\frac{3}{2}} $
$\text { R.H.S. }=\sqrt{\frac{a c e}{b d f}}=\sqrt{\frac{b k \cdot d k \cdot f k}{b d f}}=k^{\frac{3}{2}} $
$\text { L.H.S. }=\text { R.H.S. }$
Hence proved.
$\therefore a = bk , c = dk \text {, e }= fk $
$\text { L.H.S. } $
$=\left(\frac{a^2 b^2+c^2 d^2+e^2 f^2}{a b^3+c d^3+e f^3}\right)^{\frac{3}{2}} $
$=\left(\frac{b^2 k^2 \cdot b^2+d^2 k^2 \cdot d^2+f^2 k^2 \cdot f^2}{b k \cdot b^3+d k \cdot d^3+f k \cdot f^3}\right)^{\frac{3}{2}} $
$=\left[\frac{k^2\left(b^4+d^4+f^4\right)}{k\left(b^4+d^4+f^4\right)}\right]^{\frac{3}{2}} $
$=k^{\frac{3}{2}} $
$\text { R.H.S. }=\sqrt{\frac{a c e}{b d f}}=\sqrt{\frac{b k \cdot d k \cdot f k}{b d f}}=k^{\frac{3}{2}} $
$\text { L.H.S. }=\text { R.H.S. }$
Hence proved.
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