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\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(a^3-c^3\right)+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left[\left(a^3-b^3\right)+\left(b^3-c^3\right)\right]+c\left(a^3-b^3\right)\)
\(=a\left(b^3-c^3\right)-b\left(b^3-c^3\right)-b\left(a^3-b^3\right)+c\left(a^3-b^3\right)\)
\(=\left(b^3-c^3\right)\left(a-b\right)-\left(a^3-b^3\right)\left(b-c\right)\)
\(=\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-b\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[\left(b^2+bc+c^2\right)-\left(a^2+ab+b^2\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(bc+c^2-a^2-ab\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[b\left(c-a\right)+\left(c-a\right)\left(c+a\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)\)
a(b3−c3)+b(c3−a3)+c(a3−b3)
=a(b3−c3)−b(a3−c3)+c(a3−b3)
=a(b3−c3)−b[(a3−b3)+(b3−c3)]+c(a3−b3)
=a(b3−c3)−b(b3−c3)−b(a3−b3)+c(a3−b3)
=(b3−c3)(a−b)−(a3−b3)(b−c)
=(b−c)(b2+bc+c2)(a−b)−(a−b)(a2+ab+b2)(b−c)
=(a−b)(b−c)[(b2+bc+c2)−(a2+ab+b2)]
=(a−b)(b−c)(bc+c2−a2−ab)
=(a−b)(b−c)[b(c−a)+(c−a)(c+a)]
=(a−b)(b−c)(c−a)(a+b+c)
Đặt A = a + b ; B = a - b
A^3 + B^3
= (A + B)(A² - AB + B² )
= (a + b + a - b)[(a + b)² - (a + b)(a - b) + (a - b)²]
= 2a( a² + 2ab + b² - a² + b² + a² - 2ab + b² )
= 2a( a² + 3b²)
(a+b)\(^3\) - (a-b)\(^3\)
= [ (a+b) - (a-b) ] [ (a+b)\(^2\) + (a+b)(a-b) + (a-b)\(^2\) ]
= [ a+b - a+b ] [ a\(^2\) + 2ab + b\(^2\) + a\(^2\) - b\(^2\) + a\(^2\) - 2ab + b\(^2\) ]
= 2b ( 3a\(^2\) + b\(^2\) )
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)
\(=a^3+b^3+3a^2b+3ab^2+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)
\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+2abc+b^2c+bc^2\right)\)
\(=3\left(a^2b+ab^2+a^2c+ac^2+abc+abc+b^2c+bc^2\right)\)
\(=3\left[ab\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)\right]\)
\(=3\left(a+b\right)\left(ab+c^2+ac+bc\right)\)
\(=3\left(a+b\right)\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
ab(a+b)+bc(b+c)=a2b+ab2+b2c+bc2=(.................không phân tích đc nhé
zễ mà
1,2 áp dụng hằng đẳng thức
3)(2x+y)3
dùng hằng đẳng thức để phân tích:
1) \(\left(a+b\right)^3+\left(a-b\right)^3=\left[\left(a+b\right)+\left(a-b\right)\right]\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=\left(a+b+a-b\right)\left(a^2+2ab+b^2+b^2-a^2+a^2-2ab+b^2\right)\)
\(=2a\left(a^2+3b^2\right)\)
2)\(\left(a+b\right)^3-\left(a-b\right)^3=\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=\left(a+b+a-b\right)\left(a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right)\)
\(=2a\left(3a^2+b^2\right)\)
3)\(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3.\left(2x\right)^2.y+3.2x.y^2+y^3=\left(2x+y\right)^3\)