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Question Number 171204 by mokys last updated on 09/Jun/22

if : bc^2 +ca^2 +ab^2 −a^2 b−b^2 c−c^2 a=0 prove that :a = b , b = c , c = a

$${if}\::\:{bc}^{\mathrm{2}} +{ca}^{\mathrm{2}} +{ab}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}−{b}^{\mathrm{2}} {c}−{c}^{\mathrm{2}} {a}=\mathrm{0} \\ $$$${prove}\:{that}\::{a}\:=\:{b}\:,\:{b}\:=\:{c}\:,\:{c}\:=\:{a} \\ $$

Answered by som(math1967) last updated on 10/Jun/22

ab^2 −a^2 b+bc^2 −c^2 a+ca^2 −b^2 c=0 ⇒−ab(a−b)−c^2 (a−b)+c(a^2 −b^2 )=0 ⇒(a−b){−ab−c^2 +c(a+b)}=0 ⇒(a−b)(bc−c^2 −ab+ca)=0 ⇒(a−b){c(b−c)−a(b−c)}=0 ⇒(a−b)(b−c)(c−a)=0 ∴a=b,b=c,c=a

$${ab}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}+{bc}^{\mathrm{2}} −{c}^{\mathrm{2}} {a}+{ca}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}=\mathrm{0} \\ $$$$\Rightarrow−{ab}\left({a}−{b}\right)−{c}^{\mathrm{2}} \left({a}−{b}\right)+{c}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)\left\{−{ab}−{c}^{\mathrm{2}} +{c}\left({a}+{b}\right)\right\}=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)\left({bc}−{c}^{\mathrm{2}} −{ab}+{ca}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)\left\{{c}\left({b}−{c}\right)−{a}\left({b}−{c}\right)\right\}=\mathrm{0} \\ $$$$\Rightarrow\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)=\mathrm{0} \\ $$$$\therefore{a}={b},{b}={c},{c}={a} \\ $$$$ \\ $$

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