\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)
=>(1/a+1/b+1/c)^2=9
=>\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2\left(\dfrac{1}{ab}+\dfrac{1}{ac}+\dfrac{1}{bc}\right)=9\)
=>\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=9-2\cdot\dfrac{a+b+c}{abc}=9-2\cdot\dfrac{3abc}{abc}=3\)