Question
If $a, b, c$ and $d$ are in proportion, prove that:
$
\frac{a^2+a b+b^2}{a^2-a b+b^2}=\frac{c^2+c d+d^2}{c^2-c d+d^2}
$
$
\frac{a^2+a b+b^2}{a^2-a b+b^2}=\frac{c^2+c d+d^2}{c^2-c d+d^2}
$
✓
Answer
$\begin{aligned} & \because a , b , c , d \text { are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k \text { (say) } \\ & a = bk , c = dk . \\ & \text { L.H.S. }=\frac{a^2+a b+b^2}{a^2-a b+b^2} \\ & =\frac{b^2 k^2+b k \cdot b+b^2}{b^2 k^2-b k \cdot b+b^2} \\ & =\frac{b^2\left(k^2+k+1\right)}{b^2\left(k^2-k+1\right)} \\ & =\frac{k^2+k+1}{k^2-k+1} \\ & \text { R.H.S. }=\frac{c^2+c d+d^2}{c^2-c d+d^2} \\ & =\frac{d^2 k^2+d k d+d^2}{d^2 k^2-d k \cdot d+d^2} \\ & =\frac{d^2\left(k^2+k+1\right)}{d^2\left(k^2-k+1\right)} \\ & =\frac{k^2+k+1}{k^2-k+1} \\ & \therefore \text { L.H.S. = R.H.S. }\end{aligned}$
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