Ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Leftrightarrow ab+bc+ca=0\)
Ta đặt: \(\hept{\begin{cases}ab=x\\bc=y\\ca=z\end{cases}}\)
\(\Rightarrow x+y+z=0\)
Ta cần chứng minh
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\left(bc\right)^3+\left(ca\right)^3+\left(ab\right)^3=3\left(abc\right)^2\)
\(\Leftrightarrow x^3+y^3+z^3=3xyz\)
\(\Leftrightarrow\left(x^3+3xy\left(x+y\right)+y^3\right)+z^3-3xy\left(x+y\right)-3xyz=0\)
\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(z^2+y^2+2xy-yz-zx+z^2\right)-3xy\left(x+y+z\right)=0\)
\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)(đúng)
\(\RightarrowĐPCM\)