a) ( a² + b²+ c²)² - ( a² - b² - c²)²

= ( a² + b²+ c² + a² - b² - c²)( a² + b²+ c² - a² + b² + c²)

= 4a²( b² + c²)

b) ( a + b + c)² - ( a - b - c)² - 4ac

= ( a + b + c - a + b + c)( a + b + c + a - b - c) - 4ac

= 4a( b + c) - 4ac

= 4a( b + c - c)

= 4ab

Cảm ơn bn nha

nhanh lên với ak

Ta có :

a^3+b^3+c^3-3abc

=(a+b)^3+c^3-3ab(a+b) - 3abc

=(a+b+c)[(a+b)^2-(a+b)c+c^2]-3ab(a+b+c)

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

=> 2(a^3+b^3+c^3-3abc)= (a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca)

=(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]

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

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

Bổ đề : Chứng minh (a + b)² + (a - b)² = 2(a² + b²)

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

Áp dụng vào bài toán,ta có :

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

= 2[(b + c)² + a²] + 2[a² + (b - c)²] = 2[2a² + (b + c)² + (b - c)²] = 2[2a² + 2(b² + c²)] = 4(a² + b² + c²)

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

= 2[(a + b)² + (c + d)²] + 2[(a - b)² + (c - d)²] = 2[(a + b)² + (a - b)² + (c + d)² + (c - d)²]

= 2[2(a² + b²) + 2(c² + d²)] = 4(a² + b² + c² + d²)

câu a) cái khúc =2[(b+c)^2 +a^2] +2[a^2 +(b-c)^2] là răng 

ghi rõ ra dùm

a: Sửa đề: \(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc\)

\(=ab^2+ac^2+bc^2+ba^2+ca^2+cb^2+2abc\)

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

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

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

\(=\left(b+c\right)\cdot\left\lbrack a\left(a+b\right)+c\left(a+b\right)\right\rbrack=\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

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

\(=a^3-ab^2+a^2b-b^3+b^3-bc^2+b^2c-c^3+c^3-a^2c+ac^2-a^3\)

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

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

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

\(=\left(a-c\right)\left(-b^2+ba+bc-ac\right)=\left(a-c\right)\cdot\left\lbrack b\left(a-b\right)-c\left(a-b\right)\right\rbrack=\left(a-c\right)\left(a-b\right)\left(b-c\right)\)