Question

Cho a,b,c khác 0 t/m (a+b+c)^2=a^2+b^2+c^2.CMR: 1/a^3+1/b^3+1/c^3=3/abc

Answer 1

ta có: (a+b+c)² = a² + b² + c²

=> 2.(ab+ac+bc) = 0

ab + ac + bc = 0

=> 1/a + 1/b + 1/c = 0

Lại có: \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right).\)

                                                                \(=0.\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right)=0\)

=> 1/a³ + 1/b³ + 1/c³  -3/abc = 0

=> 1/a³ + 1/b³ + 1/c³ = 3/abc