(5) (a+b+c)(a+b+c) (a−b+c) (a+b−c) = {a+(b+c)} {-a+(b+c)} x {a-(b-c)}{a+(b-c)} = {-a²+(b+c)²} {a² (6-c)²) = − a² + ((b+c)² + (b_e}³} a² − ((b + c) (b-c)}²]]] == = a¹ +2(b²+c²2) a²-(6²-c²)² -a-b4-c4+2a²b²+2b²c²+2c²a² == 2 (fb-c)² x /h² a