Question Number 15888 by Tinkutara last updated on 15/Jun/17
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\Delta{ABC},\:{a}^{\mathrm{3}} \:\mathrm{cos}\:\left({B}\:−\:{C}\right)\:+ \\ $$$${b}^{\mathrm{3}} \:\mathrm{cos}\:\left({C}\:−\:{A}\right)\:+\:{c}^{\mathrm{3}} \:\mathrm{cos}\:\left({A}\:−\:{B}\right)\:=\:\mathrm{3}{abc} \\ $$
Answered by ajfour last updated on 15/Jun/17
Commented by ajfour last updated on 15/Jun/17
![In ΔABC Join B to D sucb that ∠ABD=∠C then ΔADB ∼ΔABC cos (B−C)=((ka^2 +a^2 −(b−kc)^2 )/(2ka^2 )) side AB =c =kb so k=c/b cos (B−C)=((((a^2 c^2 )/b^2 )+a^2 −(b−(c^2 /b))^2 )/((2a^2 c)/b)) cos (B−C)=((a^2 c^2 +a^2 b^2 −(b^2 −c^2 )^2 )/(2a^2 bc)) a^3 cos (B−C)=((a^2 [a^2 c^2 +a^2 b^2 −(b^2 −c^2 )^2 ])/(2abc)) similarly b^3 cos (C−A)=((b^2 [b^2 a^2 +b^2 c^2 −(c^2 −a^2 )^2 ])/(2abc)) c^3 cos (A−B)=((c^2 [c^2 b^2 +c^2 a^2 −(a^2 −b^2 )^2 ])/(2abc)) summing them up, we get Σa^3 cos (B−C)= ((6a^2 b^2 c^2 )/(2abc)) = 3abc .](https://www.tinkutara.com/question/Q15900.png)
$$\:{In}\:\Delta{ABC}\:{Join}\:{B}\:{to}\:{D}\:{sucb}\:{that} \\ $$$$\angle{ABD}=\angle{C} \\ $$$${then}\:\Delta{ADB}\:\sim\Delta{ABC} \\ $$$$\mathrm{cos}\:\left({B}−{C}\right)=\frac{{ka}^{\mathrm{2}} +{a}^{\mathrm{2}} −\left({b}−{kc}\right)^{\mathrm{2}} }{\mathrm{2}{ka}^{\mathrm{2}} } \\ $$$$\:{side}\:\:{AB}\:={c}\:={kb} \\ $$$${so}\:\:\:{k}={c}/{b}\: \\ $$$$\:\mathrm{cos}\:\left({B}−{C}\right)=\frac{\frac{{a}^{\mathrm{2}} {c}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+{a}^{\mathrm{2}} −\left({b}−\frac{{c}^{\mathrm{2}} }{{b}}\right)^{\mathrm{2}} }{\frac{\mathrm{2}{a}^{\mathrm{2}} {c}}{{b}}} \\ $$$$\:\mathrm{cos}\:\left({B}−{C}\right)=\frac{{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} {bc}} \\ $$$${a}^{\mathrm{3}} \mathrm{cos}\:\left({B}−{C}\right)=\frac{{a}^{\mathrm{2}} \left[{a}^{\mathrm{2}} {c}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} −\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)^{\mathrm{2}} \right]}{\mathrm{2}{abc}} \\ $$$${similarly} \\ $$$${b}^{\mathrm{3}} \mathrm{cos}\:\left({C}−{A}\right)=\frac{{b}^{\mathrm{2}} \left[{b}^{\mathrm{2}} {a}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}^{\mathrm{2}} −\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} \right]}{\mathrm{2}{abc}} \\ $$$${c}^{\mathrm{3}} \mathrm{cos}\:\left({A}−{B}\right)=\frac{{c}^{\mathrm{2}} \left[{c}^{\mathrm{2}} {b}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} −\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)^{\mathrm{2}} \right]}{\mathrm{2}{abc}} \\ $$$${summing}\:{them}\:{up},\:{we}\:{get} \\ $$$$\Sigma{a}^{\mathrm{3}} \mathrm{cos}\:\left({B}−{C}\right)=\:\frac{\mathrm{6}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} }{\mathrm{2}{abc}}\:=\:\mathrm{3}{abc}\:. \\ $$
Commented by Tinkutara last updated on 15/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$