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\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=\left(ab^3-a^3b\right)+\left(bc^3-ac^3\right)+\left(a^3c-b^3c\right)\)
\(=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)\)
\(=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c-abc+b^2c\right)\)
Ta có: VT=(a+b+c)3−a3−b3−c3VT=(a+b+c)3−a3−b3−c3
=[(a+b+c)3−a3]−(b3+c3)=[(a+b+c)3−a3]−(b3+c3)
=(b+c)[(a+b+c)2+(a+b+c)a+a2]−(b+c)(b2−bc+c2)=(b+c)[(a+b+c)2+(a+b+c)a+a2]−(b+c)(b2−bc+c2)
=(b+c)(3a2+3ab+3bc+3ca)=(b+c)(3a2+3ab+3bc+3ca)
=3(b+c)[a(a+b)+c(a+b)]=3(b+c)[a(a+b)+c(a+b)]
=3(a+b)(b+c)(c+a)=VP=3(a+b)(b+c)(c+a)=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(a-c)+c3(a-c-b+c)
=a3(b-c)-b3(a-c)+c3(a-c)-c3(b-c)
=(a3-c3)(b-c)-(b3-c3)(a-c)
=(a-c)(a2+ac+c2)(b-c)-(b-c)(b2+bc+c2)(a-c)
=(b-c)(a-c)(a2+ac+c2-b2-bc-c2)
=(b-c)(a-c)[(a-b)(a+b)+c(a-b)]
=(a-b)(b-c)(a-c)(a+b+c)
Ta có:\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c-b+c\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c\right)-c^3\left(b-c\right)\)
\(=\left(a^3-c^3\right)\left(b-c\right)-\left(b^3-c^3\right)\left(a-c\right)\)
\(=\left(a-c\right)\left(a^2+ac+c^2\right)\left(b-c\right)-\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-c\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(a^2+ac+c^2-b^2-bc-c^2\right)\)
\(=\left(b-c\right)\left(a-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
Lời giải:
\(a(b^3-c^3)+b(c^3-a^3)+c(a^3-b^3)\)
\(=a(b^3-c^3)-b[(b^3-c^3)+(a^3-b^3)]+c(a^3-b^3)\)
\(=(b^3-c^3)(a-b)-(a^3-b^3)(b-c)\)
\(=(b-c)(a-b)(b^2+bc+c^2)-(a-b)(b-c)(a^2+ab+b^2)\)
\(=(a-b)(b-c)(b^2+bc+c^2-a^2-ab-b^2)\)
\(=(a-b)(b-c)[(c-a)(c+a)+b(c-a)]\)
\(=(a-b)(b-c)(c-a)(c+a+b)\)