Respuesta :

stormiarmy

a+b+c=0

[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]

[a^2+b^2+c^2+2ab+2ac+2bc=0]

[a^2+b^2+c^2=-(2ab+2ac+2bc)]

[a^2+b^2+c^2=-2(ab+ac+bc)] (i)

also

[a=-b-c]

[a^2=-ab-ac] (ii)

[-c=a+b]

[-bc=ab+b^2] (iii)

adding (ii) and (iii) ,we have

[a^2-bc=b^2-ac] (iv)

devide (i) by (iv)

[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]

Otras preguntas