Algebra Questions

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

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

Answered by mrW1 last updated on 22/Jan/17

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