Ta có:

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

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

\(=-\left(c+d\right)^3+3ab\left(c+d\right)+\left(c+d\right)^3-3cd\left(c+d\right)\) (vì \(a+b=-\left(c+d\right)\))

\(=3\left(c+d\right)\left(ab-cd\right)\) 

Vậy đẳng thức được chứng minh.

...

lần sau nhé

Ta có : \(a+b+c+d=0\)

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

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

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

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

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.cd.\left(a+b\right)+3ab.\left(c+d\right)\)

\(\Leftrightarrow a^3+b^3+c^3+d^3=3.\left(c+d\right)\left(cd+ab\right)\)

Ta có : 

⇔a3+b3+c3+d3=3.cd.(a+b)+3ab.(c+d)

a+b+c=0 nên a+b=-c

a^3+b^3+c^3

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

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

=-3ab(-c)=3abc

(2x-2023)^3+(2020-x)^3+(23-x)^3=0

=>(2020-x)^3+(23-x)^3+[-(2020-x+23-x)^3]=0

=>3(2020-x)(23-x)(2x-2023)=0

=>\(x\in\left\{2020;23;\dfrac{2023}{2}\right\}\)

Chọn A

giúp mk làm với các bạn