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Question Number 181243 by Agnibhoo98 last updated on 23/Nov/22
Solve for x :  ((x − a^2 )/(b + c)) + ((x − b^2 )/(c + a)) + ((x − c^2 )/(a + b)) = 4(a + b + c)
$$\mathrm{Solve}\:\mathrm{for}\:{x}\:: \\ $$$$\frac{{x}\:−\:{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{x}\:−\:{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{x}\:−\:{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:=\:\mathrm{4}\left({a}\:+\:{b}\:+\:{c}\right) \\ $$
Answered by Frix last updated on 23/Nov/22
x=(a+b+c)^2   x−a^2 =b^2 +c^2 +2ab+2ac+2bc=  =(b+c)^2 +2ab+2ac=(b+c)^2 +2a(b+c)=  =(b+c)(2a+b+c)  the rest is plain to see
$${x}=\left({a}+{b}+{c}\right)^{\mathrm{2}} \\ $$$${x}−{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{bc}= \\ $$$$=\left({b}+{c}\right)^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{ac}=\left({b}+{c}\right)^{\mathrm{2}} +\mathrm{2}{a}\left({b}+{c}\right)= \\ $$$$=\left({b}+{c}\right)\left(\mathrm{2}{a}+{b}+{c}\right) \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{plain}\:\mathrm{to}\:\mathrm{see} \\ $$

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