Question Number 163651 by mathlove last updated on 09/Jan/22
$${if}\:\:\left({a}−\mathrm{2}{b}\right)^{\mathrm{2}} +\left({b}−\mathrm{2}{c}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${find}\:\:{volve}\:\:\frac{\left({b}+{c}−{a}\right)^{\mathrm{3}} }{{abc}}=? \\ $$
Commented by cortano1 last updated on 09/Jan/22
$$\left.\begin{matrix}{{a}=\mathrm{2}{b}\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}}{a}}\\{{b}=\mathrm{2}{c}\Rightarrow{c}=\frac{\mathrm{1}}{\mathrm{4}}{a}}\end{matrix}\right\}\:\Rightarrow\frac{−\frac{\mathrm{1}}{\mathrm{64}}{a}^{\mathrm{3}} }{\frac{\mathrm{1}}{\mathrm{8}}{a}^{\mathrm{3}} }\:=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Jan/22
$${a}=\mathrm{2}{b}\:\wedge{b}=\mathrm{2}{c}\Rightarrow{a}=\mathrm{2}\left(\mathrm{2}{c}\right)=\mathrm{4}{c} \\ $$$$\:\frac{\left({b}+{c}−{a}\right)^{\mathrm{3}} }{{abc}}=\frac{\left(\mathrm{2}{c}+{c}−\mathrm{4}{c}\right)^{\mathrm{3}} }{\mathrm{4}{c}.\mathrm{2}{c}.{c}}=\frac{−{c}^{\mathrm{3}} }{\mathrm{8}{c}^{\mathrm{3}} }=−\frac{\mathrm{1}}{\mathrm{8}} \\ $$