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Question Number 142025 by mnjuly1970 last updated on 25/May/21

Answered by som(math1967) last updated on 25/May/21

(a+b+c)^2 =1 2(ab+bc+ca)=1−2 ab+bc+ca=−(1/2) a^3 +b^3 +c^3 −3abc+3abc=3 (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) +3abc=3 1×(2+(1/2))+3abc=3 3abc=3−2(1/2)=(1/2) ∴abc=(1/6) ∴(c)abc=(1/6)

$$\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{2}\left(\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}\right)=\mathrm{1}−\mathrm{2} \\ $$$$\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{{a}}^{\mathrm{3}} +\boldsymbol{{b}}^{\mathrm{3}} +\boldsymbol{{c}}^{\mathrm{3}} −\mathrm{3}\boldsymbol{{abc}}+\mathrm{3}\boldsymbol{{abc}}=\mathrm{3} \\ $$$$\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)\left(\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} −\boldsymbol{{ab}}−\boldsymbol{{bc}}−\boldsymbol{{ca}}\right) \\ $$$$\:\:\:\:+\mathrm{3}\boldsymbol{{abc}}=\mathrm{3} \\ $$$$\mathrm{1}×\left(\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{3}\boldsymbol{{abc}}=\mathrm{3} \\ $$$$\mathrm{3}\boldsymbol{{abc}}=\mathrm{3}−\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\therefore\boldsymbol{{abc}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\therefore\left(\boldsymbol{{c}}\right)\boldsymbol{{abc}}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$

Commented by mnjuly1970 last updated on 25/May/21

thank you sir Som...

$${thank}\:{you}\:{sir}\:{Som}... \\ $$

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