Cho a+b+c+d=0,CMR: a³+b³+c³+d³=3(b+c)(ad-bc)

$\begin{array}{l}a^3 + b^3 + c^3 + d^3\\ = (a^3 + d^3) + (b^3 + c^3)\\ = (a+d)^3 – 3ad(a+ d) + (b+c)^3 – 3bc(b+c)\\ = (a + b + c + d)[(a+d)^2 – (a+d)(b+c) + (b+c)^2] – 3ad(a+d) – 3bc(b+c)\qquad (*)\\ Ta \,\,có:\\ a +b + c + d = 0\\ \Rightarrow \begin{cases}(a + b + c + d)[(a+d)^2 – (a+d)(b+c) + (b+c)^2] = 0\\a +d = – b – c\end{cases}\\ \text{Thay vào $(*)$ ta được:}\\ a^3 + b^3 + c^3 + d^3\\ = -3ad(-b-c) – 3bc(b+c)\\ = 3ad(b+c) – 3bc(b+c)\\ = 3(b+c)(ad – bc) \end{array}$