Question-220340

Question Number 220340 by Noorzai last updated on 11/May/25

Answered by som(math1967) last updated on 11/May/25

b^2 a−a^2 b+c^2 b−c^2 a+b^2 c−a^2 c =ba(b−a)+c^2 (b−a)+c(b^2 −a^2 ) =ba(b−a)+c^2 (b−a)+c(b−a)(b+a) =(b−a)(ba+c^2 +bc+ca) =(b−a){a(b+c)+c(b+c)} =(b−a)(b+c)(a+c)

$$\:{b}^{\mathrm{2}} {a}−{a}^{\mathrm{2}} {b}+{c}^{\mathrm{2}} {b}−{c}^{\mathrm{2}} {a}+{b}^{\mathrm{2}} {c}−{a}^{\mathrm{2}} {c} \\ $$$$={ba}\left({b}−{a}\right)+{c}^{\mathrm{2}} \left({b}−{a}\right)+{c}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$={ba}\left({b}−{a}\right)+{c}^{\mathrm{2}} \left({b}−{a}\right)+{c}\left({b}−{a}\right)\left({b}+{a}\right) \\ $$$$=\left({b}−{a}\right)\left({ba}+{c}^{\mathrm{2}} +{bc}+{ca}\right) \\ $$$$=\left({b}−{a}\right)\left\{{a}\left({b}+{c}\right)+{c}\left({b}+{c}\right)\right\} \\ $$$$=\left({b}−{a}\right)\left({b}+{c}\right)\left({a}+{c}\right) \\ $$

Answered by efronzo1 last updated on 11/May/25

a(b^2 −c^2 )+b(c^2 −a^2 )+c(b^2 −a^2 ) = a(b+c)(b−c)+b(c+a)(c−a)+c(b−a)(b+a)

$$\:{a}\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)+{b}\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+{c}\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\:=\:{a}\left({b}+{c}\right)\left({b}−{c}\right)+{b}\left({c}+{a}\right)\left({c}−{a}\right)+{c}\left({b}−{a}\right)\left({b}+{a}\right) \\ $$$$\: \\ $$

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Question-220340

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