if (a−2b)^2 +(b−2c)^2 =0 find volve (((b+c−a)^3 )/(abc))=?


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Question Number 163651 by mathlove last updated on 09/Jan/22

$i f {(a - 2 b)}^{2} + {(b - 2 c)}^{2} = 0$ $f i n d v o l v e \frac{{(b + c - a)}^{3}}{a b c} = ?$
Commented by cortano1 last updated on 09/Jan/22
${: ((a=2b⇒b=(1/2)a)),((b=2c⇒c=(1/4)a)) } ⇒((−(1/(64))a^3 )/((1/8)a^3 )) =−(1/8)$
$\begin{matrix} a = 2 b \Rightarrow b = \frac{1}{2} a \\ b = 2 c \Rightarrow c = \frac{1}{4} a \end{matrix}} \Rightarrow \frac{- \frac{1}{64} a^{3}}{\frac{1}{8} a^{3}} = - \frac{1}{8}$
	Answered by Rasheed.Sindhi last updated on 09/Jan/22

	$a = 2 b \land b = 2 c \Rightarrow a = 2 (2 c) = 4 c$ $\frac{{(b + c - a)}^{3}}{a b c} = \frac{{(2 c + c - 4 c)}^{3}}{4 c . 2 c . c} = \frac{- c^{3}}{8 c^{3}} = - \frac{1}{8}$
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