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

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

$${if}\:\:\left({a}−\mathrm{2}{b}\right)^{\mathrm{2}} +\left({b}−\mathrm{2}{c}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:\:{volve}\:\:\frac{\left({b}+{c}−{a}\right)^{\mathrm{3}} }{{abc}}=? \\ $$

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)

$$\left.\begin{matrix}{{a}=\mathrm{2}{b}\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}}{a}}\\{{b}=\mathrm{2}{c}\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{4}}{a}}\end{matrix}\right\}\:\Rightarrow\frac{−\frac{\mathrm{1}}{\mathrm{64}}{a}^{\mathrm{3}} }{\frac{\mathrm{1}}{\mathrm{8}}{a}^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$

Answered by Rasheed.Sindhi last updated on 09/Jan/22

a=2b ∧b=2c⇒a=2(2c)=4c   (((b+c−a)^3 )/(abc))=(((2c+c−4c)^3 )/(4c.2c.c))=((−c^3 )/(8c^3 ))=−(1/8)

$${a}=\mathrm{2}{b}\:\wedge{b}=\mathrm{2}{c}\Rightarrow{a}=\mathrm{2}\left(\mathrm{2}{c}\right)=\mathrm{4}{c} \\ $$$$\:\frac{\left({b}+{c}−{a}\right)^{\mathrm{3}} }{{abc}}=\frac{\left(\mathrm{2}{c}+{c}−\mathrm{4}{c}\right)^{\mathrm{3}} }{\mathrm{4}{c}.\mathrm{2}{c}.{c}}=\frac{−{c}^{\mathrm{3}} }{\mathrm{8}{c}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$

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