Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Leftrightarrow x+y+z=a+b+c\)

Do đó \(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(\Leftrightarrow A=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ \Leftrightarrow A=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Leftrightarrow A=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\\ \Leftrightarrow A=3\cdot2b\cdot2c\cdot2a=24abc\)

Ta có: $VT={\left(a+b+c\right)}^3-a^3-b^3-c^3$

$=\left[{\left(a+b+c\right)}^3-a^3\right]-\left(b^3+c^3\right)$

$=\left(b+c\right)\left[{\left(a+b+c\right)}^2+\left(a+b+c\right)a+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)$

$=\left(b+c\right)\left(3a^2+3ab+3bc+3ca\right)$

$=3\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]$

$=3\left(a+b\right)\left(b+c\right)\left(c+a\right)=VP$ (Đpcm)

Thật ra mình làm theo đề thấy nó đáng ra phải là chứng minh chứ ko phải phân tích . chúc học tốt!

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

Có vẻ nhìn hơi rối mắt bạn thông cảm nha

a3(b−c)+b3(c−a)+c3(a−b)

=a3b−a3c+b3c−b3a+c3(a−b)

=(a3b−b3a)−(a3c−b3c)+c3(a−b)

=ab(a2−b2)−c(a3−b3)+c3(a−b)

=ab(a−b)(a+b)−c(a−b)(a2+ab+b2)+c3(a−b)

=(a−b)[ab(a+b)−c(a2+ab+b2)+c3]

=(a−b)(a2b+ab2−a2c−abc−b2c+c3)

=(a−b)[(a2b−a2c)+(ab2−abc)−(b2c−c3)]

=(a−b)[a2(b−c)+ab(b−c)−c(b2−c2)]

=(a−b)[a2(b−c)+ab(b−c)−c(b−c)(b+c)]

=(a−b)(b−c)[a2+ab−c(b+c)]

=(a−b)(b−c)(a2+ab−bc−c2)

=(a−b)(b−c)[(a−c)(a+c)+b(a−c)]

=(a−b)(b−c)(a−c)(a+b+c)

ko bt đâu nhá!

Lời giải:

\(a(b^3-c^3)+b(c^3-a^3)+c(a^3-b^3)\)

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

\(=(b^3-c^3)(a-b)-(a^3-b^3)(b-c)\)

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

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

\(=(a-b)(b-c)[(c-a)(c+a)+b(c-a)]\)

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