Ta có :

\(M=a^3+b^3+c\left(a^2+b^2\right)-abc\)

\(M=a^3+b^3+a^2c+b^2c-abc\)

\(=\left(a^3+a^2c\right)+\left(b^3+b^2c\right)-abc\)

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

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

\(=-ab\left(a+b+c\right)=0\)

Ta có: \(a+b+c=0\)

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

\(M=a^3+b^3+c.\left(a^2+b^2\right)-abc\)

\(M=a^3+b^3+ca^2+cb^2-abc\)

\(M=a^2.\left(a+c\right)+b^2.\left(b+c\right)-abc\)

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

\(M=-a^2b-b^2a\)

\(M=-ab.\left(a+b\right)\)

\(M=-ab.\left(-c\right)\)

\(M=abc\)

Tham khảo nhé~

Ta có a+b+c=0 <=> a+b=-c

M= (a+b)(a²-ab+b²)+a²c+b²c-abc

=-ca²+abc-cb²+a²c+b²c-abc

=0

Chúc học tốt!!!!

\(A^3+B^3+A^2C+B^2C-ABC\)

\(=\left(A+B\right)\left(A^2-AB+B^2\right)+C\left(A^2-AB+B^2\right)\)

\(=\left(A^2-AB+B^2\right)\left(A+B+C\right)\)

\(=\left(A^2-AB+B^2\right).0\)

\(=o\)

là 0 chứ rút gọn gì nữa

Ta có: \(M=a^3+b^3+c\left(a^2+b^2\right)-abc\)

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

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

\(M=0.\left(a^2+b^2-ab\right)\)

\(M=0\)

Vậy \(M=0\)

\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{\left(a-b\right)\left(a+b\right)-c^2}=\dfrac{a^2}{\left(a-b\right)\left(-c\right)-c^2}=\dfrac{a^2}{c\left(b-a-c\right)}=\dfrac{a^2}{2bc}\\ \Leftrightarrow M=\sum\dfrac{a^2}{a^2-b^2-c^2}=\sum\dfrac{a^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}\\ \Leftrightarrow M=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)}{2abc}=0\)

tự làm nhé,dễ lắm

bài này khó đấy

Bài này dễ mà

\(a+b=-c\Leftrightarrow\left(a+b\right)^3=-c^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Leftrightarrow a^3+b^3+c^3=-3ab\left(a+b\right)=3abc\)

\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)

A, B đâu em?