Ta có: \(a+b+c+d=0\Rightarrow a+b=-c-d=-\left(c+d\right)\)
Do đó: \(\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3=-c^3-3c^2d-3cd^2-d^3\)
\(\Leftrightarrow a^3+3ab\left(a+b\right)+b^3=-c^3-3cd\left(c+d\right)-d^3\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3cd\left(c+d\right)-3ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3cd\left(a+b\right)-3ab\left(a+b\right)\) (vì \(a+b=-\left(c+d\right)\))
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(a+b\right)\left(cd-ab\right)\)
Suy ra: \(a^3+b^3+c^3+d^3-3\left(a+b\right)\left(cd-ab\right)=0\)
Vậy \(a^3+b^3+c^3+d^3-3\left(a+b\right)\left(cd-ab\right)=0\) với \(a+b+c+d=0\)
Các cao nhân giúp em ạ
