Bài 1: CMR
a/ 2*(a^3+ b^3+ c^3- 3abc)=(a+b+c)*((a-b)^2+(b-c)^2+(c-a)^2)
b/ (a+b)*(b+c)*(c+a)+4abc=c*(a+b)^2+a*(b+c)^2+b*(c+a)^2
c/ (a+b+c)^3=a^3+b^3+c^3+3*(a+b)*(b+c)*(c+a)
Bài 2: Cho a+b+c=4m.CMR:
a/ 2ab+ a^2+ b^2- c^2=16m^2- 8mc
b/ (a+b-c/2)^2+(a-b+c/2)^2+(b+c-a/2)^2=a^2+b^2+c^2-4m^2
cmr
c) (a+b+c)3 -a 3 -b 3 -c 3=3(a+b)(b+c)(c+a)
d) a3+b3+c3 -3abc=(a+b+c)(a2+b2 +c2 -ab-bc-ca)
e) (a+b+c)3 -(b+c-a)3 -(a+c-b) 3 -(a+b-c)3=24abc
Phân tích thành nhân tử
1, a(b-c)3+b(c- a)3+c(a- b)
2, a^4(b-c)+b^4(c-a)+c^4(a-b)
3, bc(a+d)(b-c)-ac(b+d)(a-c)+ab(c+d)(a-b)
4, (a+b+c)^3-(a+b-c)^3-(b+c-a)^3-(c+a-b)^3
5, (b-c)^3+(c-a)^3+(a-b)^3
chứng minh hàng đẳng thức:
a) (a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b) (b+c) (c+a)
b) (a+b+c) ^3 - a^3 - b^3 -c^3 = 3(a+b)(b+c)(c+a)
Giúp mình với, mình cần rất gấp
CMR
a, (a+b)(b+c)(c+a)+4abc=c(a+b)^2+a(b+c)^2+b(c+a)^2
b, (a^3+b^3+c^3)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)
thank!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Cho a+b+c+d=0
a) Chứng minh a^3+b^3+c^3+d^3=3(ab-cd)(c+d)
b)Chứng minh (a+b+c+)^3=a^3 + b^3 + c^3+3(a+b)(b+c)(c+a)
c)Cho c-a=b+d. Chứng Minh a^3+b^3-c^3+d^3=3(d-c)(ab+cd)
Rút gọn biểu thức:
a) a+b+c)^3 - (b+c-a) ^3 - (a+c-b)^3 - (a+b-c)^3
b)(a+b)^3 + (b+c)^3 + (c+a) -3(a+b)(b+c)(c+a)
Giúp mình với nhé
(1) (a+b+c)2=a2+b2+c2+2ab+2bc+2ac(a+b+c)2=a2+b2+c2+2ab+2bc+2ac
(2) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) (a−b−c)2=a2+b2+c2−2ab−2ac+2bc(a−b−c)2=a2+b2+c2−2ab−2ac+2bc
(4) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) a3−b3=(a−b)3+3ab(a−b)a3−b3=(a−b)3+3ab(a−b)
(6) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ac)
(8) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2(a+b)(b+c)(c+a)−8abc=a(b−c)2+b(c−a)2+c(a−b)2
(10) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33ab2+bc2+ca2−a2b−b2c−c2a=(a−b)3+(b−c)3+(c−a)33
(12)ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Chứng minh giùm mik hằng đẳng thức kia vs