Question

cho a,b,c khác 0 thỏa \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) chứng minh rằng \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Answer 1

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

=> x+y+z=0

Có \(x^3+y^3+z^3-3xyz\)

=\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)

=\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2-3xy\right]\)

=0( do x+y+z=0)

=> \(x^3+y^3+z^3=3xyz\)

<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)