Question

Cho\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

Chứng minh:\(\dfrac{1}{a^3+b^3+c^3}=\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\)

Answer 1

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\)

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

=> (bc+ac+ab)(a+b+c)=abc

=> abc+b²c+bc²+a²c+abc+ac²+a²b+ab²+abc=abc

=>abc+b²c+bc²+a²c+abc+ac²+a²c+ab²+abc-abc=0

=>(a²c+2abc+b²c)+(a²b+ab²)+(ac²+bc²)=0

=>c(a+b)²+ab(a+b)+c²(a+b)=0

=>(a+b)[c(a+b)+ab+c²]=0

=>(a+b)(ac+bc+ab+c²)=0

=>(a+b)[a(c+b)+c(b+c)]=0

=>(a+b)(c+b)(a+c)=0

=> a+b=0, c+b=0, a+c=0

nếu a+b=0=>a=-b

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

và \(\dfrac{1}{a^3+b^3+c^3}=\dfrac{1}{-b^3+b^3+c^3}=\dfrac{1}{c^3}\) (2)

từ (1) và (2) suy ra đpcm