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\(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)
\(=\left(a+b\right)^3+3\cdot c\cdot\left(a+b\right)^2+3\cdot c^2\left(a+b\right)+c^3\)
\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2+c^3\)
\(=a^3+b^3+c^3+3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
1) a3+b3+c3-3abc = (a+b)3-3ab(a+b)+c3-3abc
= (a+b+c)(a2+2ab+b2-ab-ac+c2) -3ab(a+b+c)
= (a+b+c)( a2+b2+c2-ab-bc-ca)
Bài 2:
Ta có: \(a+b+c=0\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)
\(\Rightarrow a^3+b^3+3ab.\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+3ab.\left(-c\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
(Còn nhiều cách nữa ,mình làm 1 cách nhé)
Ta có
a3+b3+c3=a3+3ab(a+b)+b3+c3-3ab(a+b)
=(a+b)3+c3-3ab(a+b)
=(a+b+c)[(a+b)2-(a+b)c+c2 ]-3ab(a+b+c)+3abc
=(a+b+c)(a2+b2+c2+2ab-ac-bc-3ab)+3abc
=(a+b+c)(a2+b2+c2-ab-bc-ca)+3abc
Tớ chỉ phân tích đc như vậy thôi !!!
\(\left[a^2+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]-\left(a+b+c\right)^3\)
\(=\left(a^3+b^3+c^3+\left(3a+3b\right)\cdot\left(b+c\right)\cdot\left(c+a\right)\right)-\\ \left(\left(a+b\right)^2+3c\cdot\left(a+b\right)^2+3\left(a+b\right)\cdot c^2+c^3\right)\)
\(=\left(a^3+b^3+c^3+\left(3ab+3ac+3b^2+3bc\right)\cdot\left(c+a\right)\right)-\\ \left(a^2+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2+c^3\right)\)
\(=\left(a^3+b^3+c^3+3abc+3a^2b+3ac^2+3a^2c+3ab^2+3bc^2\cdot3bc^2+3abc\right)-\\ \left(a^3+3a^2b+3ab^2+b^3+3a^2c+6abc+3b^2c+3ac^2+3bc^2+c^3\right)\)
\(=\left(a^3+b^3+c^3+6abc+3a^2b+3ac^2+3a^2c+3b^2c+3ab^2+3bc^2\right)-\\ a^3-3a^2b-3ab^2-b^3-3a^2c-6abc-3b^2c-3ac^2-3bc^2-c^3\)
\(=a^3+b^3+c^3+6abc+3a^2+3ac^2+3a^2c+3ab^2+3bc^2-a^3-\\ 3a^2b-3ab^2-b^3-3a^2c-6abc-3b^2c-3ac^2-3bc^2-c^3\)
\(=\left(a^3-a^3\right)+\left(b^3-b^3\right)+\left(c^3-c^3\right)+\left(6abc-6abc\right)+\left(3a^2b-3a^2b\right)\\ +\left(3ac^2-3ac^2\right)+\left(3a^2c-3a^2c\right)+\left(3ab^2-3ab^2\right)+\left(3ab^2-3ab^2\right)+\left(3bc^2-3bc^2\right)\)
\(=0\)
=> \(\left(a+b+c\right)^3=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có :
a^3+b^3+c^3-3abc=0
<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc=0
câu 2:<=>[(a+b)^3 +c^3] -3ab.(a+b+c)=0
<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)=0
<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab)...
<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)=0
luôn đúng do a+b+c=0
câu 1:(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c3^+3(a+b)(ab+ac+bc+c2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)
CHÚC BẠN HỌC TỐT^^