Question
If $a, b, c$ and d are in proportion, prove that: $(a^4 + c^4) : (b^4 + d^4) = a^2 c^2 : b^2 d^2.$
✓
Answer
$
\begin{aligned}
& \because a, b, c, d \text { are in proportion } \\
& \frac{a}{b}=\frac{c}{d}=k(\text { say }) \\
& a=b k, c=d k . \\
& \left(a^4+c^4\right):\left(b^4+d^4\right)=a^2 c^2: b^2 d^2
\end{aligned}
$
$
\text { L.H.S. }=\frac{a^4+c^4}{b^4+d^4}
$
$
=\frac{b^4 k^4+d^4 k^4}{b^4+d^4}
$
$
\begin{aligned}
& =\frac{k^4\left(b^4+d^4\right)}{\left(b^4+d^4\right)} \\
& = k ^4
\end{aligned}
$
$
\text { R.H.S. }=\frac{a^2 c^2}{b^2 d^2}
$
$
\begin{aligned}
& =\frac{k^2 b^2 \cdot k^2 d^2}{b^2 \cdot d^2} \\
& = k ^4
\end{aligned}
$Hence $ \text{L.H.S. = R.H.S.}$
\begin{aligned}
& \because a, b, c, d \text { are in proportion } \\
& \frac{a}{b}=\frac{c}{d}=k(\text { say }) \\
& a=b k, c=d k . \\
& \left(a^4+c^4\right):\left(b^4+d^4\right)=a^2 c^2: b^2 d^2
\end{aligned}
$
$
\text { L.H.S. }=\frac{a^4+c^4}{b^4+d^4}
$
$
=\frac{b^4 k^4+d^4 k^4}{b^4+d^4}
$
$
\begin{aligned}
& =\frac{k^4\left(b^4+d^4\right)}{\left(b^4+d^4\right)} \\
& = k ^4
\end{aligned}
$
$
\text { R.H.S. }=\frac{a^2 c^2}{b^2 d^2}
$
$
\begin{aligned}
& =\frac{k^2 b^2 \cdot k^2 d^2}{b^2 \cdot d^2} \\
& = k ^4
\end{aligned}
$Hence $ \text{L.H.S. = R.H.S.}$
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