\(T=\left(b+c-a\right)\left(a+c-b\right)+\left(a+c-b\right)\left(a+b-c\right)+\left(a+b-c\right)\left(b+c-a\right)\)

   \(=c^2-\left(a-b\right)^2+a^2-\left(b-c\right)^2+b^2-\left(a-c\right)^2\)

\(=\left(a^2+b^2+c^2\right)-\left(a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\right)\)

\(=2\left(ab+bc+ca\right)-\left(a^2+b^2+c^2\right)\)?????

\(\text{(S-2b)(S-2c)}=\text{(}a-b+c)\left(a+b-c\right)=a^2-\left(b-c\right)^2=a^2-b^2-c^2+2bc\left(1\right)\)

Hoán vị vòng b->c->a->b, ta được :

\(\text{(S-2c)(S-2a)}=b^2-c^2-a^2+2ca\left(2\right)\)

\(\text{(S-2a)(S-2b)}=c^2-a^2-b^2+2ab\left(3\right)\)

Từ (1),(2),(3) ta suy ra kết quả của tổng là :

\(-a^2-b^2-c^2+2ab+2bc+2ca\)

Ta có: \(S=a+b+c\left(1\right)\)

Thay \(\left(1\right)\)vào ta được:

\(\left(S-2b\right).\left(S-2c\right)=\left(a+b+c-2b\right).\)\(\left(a+b+c-2c\right)\)

                                       \(=\left(a-b+c\right).\left(a+b-c\right)\)

                                       \(=a^2+ab-ac-ba-b^2+bc+ca+cb-c^2\)

                                       \(=a^2-b^2-c^2+2.bc\left(2\right)\)

Tương tự, ta được:

\(\left(S-2c\right).\left(S-2a\right)=b^2-c^2-a^2+2.ca\left(3\right)\)

\(\left(S-2a\right).\left(S-2b\right)=c^2-a^2-b^2+2.ab\left(4\right)\)

Từ \(\left(2\right);\left(3\right);\left(4\right)\Rightarrow\)Tổng bằng:

\(a^2-b^2-c^2+2bc+b^2-c^2-a^2+2ca+c^2-a^2\)\(-b^2+2ab\)

\(=2ab+2bc+2ca-a^2-b^2-c^2\)

Vậy tổng trên \(=2ab+2bc+2ca-a^2-b^2-c^2.\)

Thay S=a+b+c vào biểu thức ta được:

(a+b+c-2b)(a+b+c-2c)+(a+b+c-2c)(a+b+c-2a)+(a+b+c-2a)(a+b+c-2b)

=(a-b+c)(a+b-c)+(b-c+a)(b+c-a)+(c-a+b)(c+a-b)

=a²-(b-c)²+b²-(c-a)²+c²-(a-b)²

=a²-b²+2bc-c²+b²-c²+2ac-a²+c²-a²+2ab-b²

=-a²-b²-c²+2ab+2bc+2ca

Trả lời giúp bạn nè:
VT = S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b)
= S((S - 2b)(S -2c) + (S-2c)(S - 2a) + (S - 2a)(S - 2b) )
= S ( S² -2cS -2bS + 4bc + S² - 2aS - 2cS +4ac + S² -2bS -2aS +4ab )
= S ( 3S² - 4cS -4bS - 4aS + 4bc + 4ac + 4ab)
= 3S³ - 4cS² - 3bS² - 4aS² + 4bcS + 4acS + 4abS
= S³ + S³ + S³ - 4cS² - 3bS² - 4aS² + 4bcS + 4acS + 4abS
= S² (S -4c ) + S² (S -4b ) + S² (S -4a )
= S² ( S -4c + S - 4b + S - 4a)
= S² (3S - 4(c + b + a)
= S² (3S - 4S)
= 3S³ - 4S³
= -S³ ( 1 )

VP = (S - 2a)(S - 2b)(S - 2c) + 8abc
= (S² -2bS -2aS + 4ab)(S - 2c) + 8abc
= S³ - 2cS² - 2bS² + 4bcS - 2aS² + 4acS + 4abS - 8abc + 8abc
= S³ - 2cS² - 2bS² - 2aS² + 4bcS + 4acS + 4abS
= S² (S -2c ) - S² (2b + 2a )
= S² ( S - 2c - 2b - 2a )
= S² ( S - 2( c + b + a))
= S³ - 2S³
= -S³ ( 2 )
Từ (1) và (2) suy ra :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S - 2c) + 8abc

thank you!!

Trả lời giúp bạn nè:
VT = S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b)
= S((S - 2b)(S -2c) + (S-2c)(S - 2a) + (S - 2a)(S - 2b) )
= S ( S2 -2cS -2bS + 4bc + S2 - 2aS - 2cS +4ac + S2 -2bS -2aS +4ab )
= S ( 3S2 - 4cS -4bS - 4aS + 4bc + 4ac + 4ab)
= 3S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S3 + S3 + S3 - 4cS2 - 3bS2 - 4aS2 + 4bcS + 4acS + 4abS
= S2 (S -4c ) + S2 (S -4b ) + S2 (S -4a )
= S2 ( S -4c + S - 4b + S - 4a)
= S2 (3S - 4(c + b + a)
= S2 (3S - 4S)
= 3S3 - 4S3
= -S3 ( 1 )

VP = (S - 2a)(S - 2b)(S - 2c) + 8abc
= (S2 -2bS -2aS + 4ab)(S - 2c) + 8abc
= S3 - 2cS2 - 2bS2 + 4bcS - 2aS2 + 4acS + 4abS - 8abc + 8abc
= S3 - 2cS2 - 2bS2 - 2aS2 + 4bcS + 4acS + 4abS
= S2 (S -2c ) - S2 (2b + 2a )
= S2 ( S - 2c - 2b - 2a )
= S2 ( S - 2( c + b + a))
= S3 - 2S3
= -S3 ( 2 )
Từ (1) và (2) suy ra :
S(S - 2b)(S -2c) + S(S-2c)(S - 2a) + S(S - 2a)(S - 2b) = (S - 2a)(S - 2b)(S - 2c) + 8abc

a)\(x^3+y^3+z^3-3xyz\\ \left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-\left[3xyz+3xy\left(x+y\right)\right]\\=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right] \\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+x^2-xy-xz-yz\right)\)