Question

Cho 3 số a, b, c khác 0 thoả:

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

Chứng minh: \(a^3+b^3+c^3⋮3\)

Answer 1

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

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{a+b+c}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)

\(\Leftrightarrow a+b+c=0\) \(\Leftrightarrow a+b=-c\)

\(\Leftrightarrow\left(a+b\right)^3=-c^3\Rightarrow\left(a+b\right)^3+c^3=0\)

Ta có:

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

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

\(=-3ab\left(a+b\right)=3abc⋮3\) (đpcm)