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Roxyy70
a fost răspuns

Aratati ca (x+3)^3-x-3 = (x+2) (x+3) (x+4), oricare ar fi x numar real.

Răspuns :

Nicumavro
(x+3)^3-x-3 = (x+2) (x+3) (x+4)
(x+3)[(x+3)^2-(x+3)]-(x+2) (x+3) (x+4)=0
(x+3)[x^2+6x+9-1]-(x+2) (x+3) (x+4)=0
(x+3)[x^2+6x+8-(x+2)(x+4)]=0
(x+3)[x^2+6x+8-(x^2+6x+8)]=0
(x+3)*0=0 indiferent de x real, ceea ce trebuia demonstrat!

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