Răspuns :
comun ai (a+b) si (a+c)
deci avem:
(a+b)x(a+c)x[(b+c)-(b-c)]=(a+b)x(a+c)x(b+c-b+c)=(a+b)x(a+c)x2c=2x(a+b)x(a+c)
deci avem:
(a+b)x(a+c)x[(b+c)-(b-c)]=(a+b)x(a+c)x(b+c-b+c)=(a+b)x(a+c)x2c=2x(a+b)x(a+c)