\(=>\left(\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{c}\right)^2=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)

Phân tích vế trái ta được ( hằng đẳng thức) :>

\(=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2+\dfrac{2}{ac}+\dfrac{2}{bc}+\left(\dfrac{1}{c}\right)^2\)

\(=\dfrac{1}{a^2}+\dfrac{2}{ab}+\dfrac{1}{b^2}+\dfrac{2}{ac}+\dfrac{2}{bc}+\dfrac{1}{c^2}\)

\(=>\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ac}+\dfrac{2}{ab}+\dfrac{2}{bc}=\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\)

\(=>2.\left(\dfrac{1}{ac}+\dfrac{1}{ab}+\dfrac{1}{bc}\right)=0\)

\(=>\dfrac{b}{abc}+\dfrac{c}{abc}+\dfrac{a}{abc}=0\)

\(=>a+b+c=0.abc=0\)

\(=>a+b=-c\)

\(=>-\left(a+b\right)=c\)

Thay vào ta có:

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

\(=-3a^2b-3ab^2=3\left(-a^2b-ab^2\right)⋮3\)

CHÚC BẠN HỌC TỐT NHA-