Question

Cho a, b , c ≠ 0 và (a + b + c)² = a² + b² + c²

Tính M = \(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\)

Answer 1

Lời giải:
\((a+b+c)^2=a^2+b^2+c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2(ab+bc+ac)=a^2+b^2+c^2\)

\(\Leftrightarrow ab+bc+ac=0\)

\(\Rightarrow ab+bc=-ac\). Từ đây suy ra:

\(M=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{(ab)^3+(bc)^3+(ca)^3}{(abc)^3}\)

\(=\frac{(ab)^3+(bc)^3+3(ab)^2(bc)+3(ab)(bc)^2-3(ab)^2(bc)-3(ab)(bc)^2+(ca)^3}{(abc)^3}\)

\(=\frac{(ab+bc)^3-3ab^2c(ab+bc)+(ca)^3}{(abc)^3}\)

\(=\frac{(-ca)^3-3ab^2c(-ca)+(ca)^3}{(abc)^3}\)

\(=\frac{3a^2b^2c^2}{(abc)^3}=\frac{3}{abc}\)