Question

Demonstrati ca expresia A = bc³+a³c+ab³-a³b-b³c-ac³ este egala cu (a-b)(b-c)(c-a)(a+b+c).

Imi trebuie rezolvarea de la stanga spre dreapta, adica bc³+a³c+ab³-a³b-b³c-ac³ sa fie egal dupa mai multe etape cu (a-b)(b-c)(c-a)(a+b+c).

Multumesc in avans!

Answer 1

$\it \underline{bc^3} +\underline{\underline{a^3c}} +\underline{\underline{\underline{ab^3-a^3b}}} -\underline{\underline{b^3c}}-\underline{ac^3}=-c^3(a-b) +c(a^3-b^3)-ab(a^2-b^2)=\\ \\ =-c^3(a-b)+c(a-b)(a^2+ab+b^2)-ab(a-b)(a+b)=\\ \\=(a-b)(\underline{-c^3}+\underline{\underline{a^2c}}+\underline{\underline{\underline{abc}}}+\underline{b^2c}-\underline{\underline{a^2b}}-\underline{\underline{\underline{ab^2}}}) =\\ \\ =(a-b)[c(b^2-c^2)-a^2(b-c)-ab(b-c)]=\\ \\ =(a-b)[c(b-c)(b+c)-a^2(b-c)-ab(b-c)]=$

$\it=(a-b)(b-c)(\underline{bc}+\underline{\underline{c^2-a^2}}-\underline{ab})=\\ \\ =(a-b)(b-c)[b(c-a)+(c-a)(c+a)]=(a-b)(b-c)(c-a)(b+c+a)=\\ \\ =(a-b)(b-c)(c-a)(a+b+c)$