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Question Number 16039 by Tinkutara last updated on 17/Jun/17

Commented by RasheedSoomro last updated on 17/Jun/17

c^2 =a^2 +b^2 −2ab cos 60°  c^2 =a^2 +b^2 −ab((1/2))=a^2 −ab+b^2   Now,  (1/(a+c))−(3/(a+b+c))+(1/(b+c))=((a^2 −ab+b^2 −c^2 )/((a+c)(a+b+c)(b+c)))           =((c^2 −c^2 )/((a+c)(a+b+c)(b+c)))=0

$$\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{2ab}\:\mathrm{cos}\:\mathrm{60}° \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{ab}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} \\ $$$$\mathrm{Now},\:\:\frac{\mathrm{1}}{\mathrm{a}+\mathrm{c}}−\frac{\mathrm{3}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}+\frac{\mathrm{1}}{\mathrm{b}+\mathrm{c}}=\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{ab}+\mathrm{b}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} }{\left(\mathrm{a}+\mathrm{c}\right)\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} }{\left(\mathrm{a}+\mathrm{c}\right)\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}+\mathrm{c}\right)}=\mathrm{0} \\ $$

Commented by Tinkutara last updated on 17/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

Commented by ajfour last updated on 17/Jun/17

 Sir, you make it so easy and   short, very good sir.

$$\:{Sir},\:{you}\:{make}\:{it}\:{so}\:{easy}\:{and} \\ $$$$\:{short},\:{very}\:{good}\:{sir}. \\ $$

Commented by RasheedSoomro last updated on 17/Jun/17

Thanks for encouraging!

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{encouraging}! \\ $$

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