Question
If $\frac{a}{c}=\frac{c}{d}=\frac{c}{f}$ prove that : $b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f}\right]^3$ $=27(a+b)(c+d)(e+f)$
✓
Answer
$\begin{aligned} & \frac{a}{c}=\frac{c}{d}=\frac{c}{f}= k \text { (say) } \\ & \therefore a = bk _{ l } c = dk _{ k } e = fk \\ & \text { L.H.S. }=b d f\left[\frac{a+b}{b}+\frac{c+d}{d}+\frac{c+f}{f}\right]^3 \\ & =b d f\left[\frac{b k+b}{b}+\frac{d k+d}{d}+\frac{f k+f}{f}\right]^3 \\ & =b d f\left[\frac{b(k+1)}{b}+\frac{d(k+1)}{d}+\frac{f(k+1)}{f}\right]^3 \\ & =\operatorname{bdf}( k +1+ k +1+ k +1)^3 \\ & =\operatorname{bdf}(3 k +3)^3=27 bdf ( k +1)^3 \\ & \text { R.H.S. }=27(a+b)(c+d)(e+f) \\ & =27(b k+b)(d k+d)(f k+f) \\ & =27 b ( k +1) d ( k +1) f ( k +1) \\ & =27 bdf ( k +1)^3 \\ & \therefore \text { L.H.S. }=\text { R.H.S. } \\ & \end{aligned}$
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