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Question Number 10067 by konen last updated on 22/Jan/17

((a×b×c)/(a+b))=−(1/2) ((a×b×c)/(b+c))=(1/2) ((a×b×c)/(a+c))=1 ⇒a×b×c=?

$$\frac{\mathrm{a}×\mathrm{b}×\mathrm{c}}{\mathrm{a}+\mathrm{b}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{a}×\mathrm{b}×\mathrm{c}}{\mathrm{b}+\mathrm{c}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{a}×\mathrm{b}×\mathrm{c}}{\mathrm{a}+\mathrm{c}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}×\mathrm{b}×\mathrm{c}=? \\ $$$$ \\ $$$$ \\ $$

Answered by mrW1 last updated on 22/Jan/17

let u=a×b×c ⇒a+b=−2u ...(i) ⇒b+c=2u ...(ii) ⇒a+c=u ...(iii) (ii)−(iii): b−a=u ...(iv) (i)+(iv): 2b=−u b=−(u/2) c=2u−b=2u+(u/2)=((5u)/2) a=−2u−b=−2u+(u/2)=−((3u)/2) u=a×b×c=(((−3u)/2))×(((−u)/2))×(((5u)/2))=((15)/8)u^3 (((15)/8)u^2 −1)u=0 u≠0 ⇒((15)/8)u^2 −1=0 u^2 =(8/(15)) ⇒a×b×c=u=±(√(8/(15)))=±2(√(2/(15)))

$${let}\:{u}={a}×{b}×{c} \\ $$$$\Rightarrow{a}+{b}=−\mathrm{2}{u}\:\:\:\:...\left({i}\right) \\ $$$$\Rightarrow{b}+{c}=\mathrm{2}{u}\:\:\:\:\:...\left({ii}\right) \\ $$$$\Rightarrow{a}+{c}={u}\:\:\:\:\:\:...\left({iii}\right) \\ $$$$\left({ii}\right)−\left({iii}\right): \\ $$$${b}−{a}={u}\:\:\:\:...\left({iv}\right) \\ $$$$\left({i}\right)+\left({iv}\right): \\ $$$$\mathrm{2}{b}=−{u} \\ $$$${b}=−\frac{{u}}{\mathrm{2}} \\ $$$${c}=\mathrm{2}{u}−{b}=\mathrm{2}{u}+\frac{{u}}{\mathrm{2}}=\frac{\mathrm{5}{u}}{\mathrm{2}} \\ $$$${a}=−\mathrm{2}{u}−{b}=−\mathrm{2}{u}+\frac{{u}}{\mathrm{2}}=−\frac{\mathrm{3}{u}}{\mathrm{2}} \\ $$$${u}={a}×{b}×{c}=\left(\frac{−\mathrm{3}{u}}{\mathrm{2}}\right)×\left(\frac{−{u}}{\mathrm{2}}\right)×\left(\frac{\mathrm{5}{u}}{\mathrm{2}}\right)=\frac{\mathrm{15}}{\mathrm{8}}{u}^{\mathrm{3}} \\ $$$$\left(\frac{\mathrm{15}}{\mathrm{8}}{u}^{\mathrm{2}} −\mathrm{1}\right){u}=\mathrm{0} \\ $$$${u}\neq\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{15}}{\mathrm{8}}{u}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${u}^{\mathrm{2}} =\frac{\mathrm{8}}{\mathrm{15}} \\ $$$$\Rightarrow{a}×{b}×{c}={u}=\pm\sqrt{\frac{\mathrm{8}}{\mathrm{15}}}=\pm\mathrm{2}\sqrt{\frac{\mathrm{2}}{\mathrm{15}}} \\ $$

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