Problem 1 Assume that for pairwise distinct real numbers a,b,c,da,b,c,da,b,c,d holds: (a2+b2−1)(a+b)=(b2+c2−1)(b+c)=(c2+d2−1)(c+d).(a^2+b^2-1)(a+b)=(b^2+c^2-1)(b+c)=(c^2+d^2-1)(c+d).(a2+b2−1)(a+b)=(b2+c2−1)(b+c)=(c2+d2−1)(c+d). Prove that a+b+c+d=0.a+b+c+d=0.a+b+c+d=0.