Question
If $a, b, c$ and $d$ are in proportion, prove that:
$
\frac{(a+c)^3}{(b+d)^3}=\frac{a(a-c)^2}{b(b-d)^2}
$
$
\frac{(a+c)^3}{(b+d)^3}=\frac{a(a-c)^2}{b(b-d)^2}
$
✓
Answer
$\begin{aligned} & \because a , b , c \text {, d are in proportion } \\ & \frac{a}{b}=\frac{c}{d}= k (\text { say) } \\ & a = bk , c = dk . \\ & \text { L.H.S. }=\frac{(a+c)^3}{(b+d)^3} \\ & =\frac{(b k+d k)^3}{(b+d)^3} \\ & =\frac{k^3(b+d)^3}{(b+d)^2} \\ & = k ^3 \\ & \text { R.H.S. }=\frac{a(a-c)^2}{b(b-d)^2} \\ & =\frac{b k(b k-d k)^2}{b(b-d)^2} \\ & =\frac{b k . k^2(b-d)^2}{b(b-d)^2} \\ & = k ^3 \\ & \therefore \text { L.H.S. }=\text { R.H.S. }\end{aligned}$
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