Question

If a + b + c = 3, a2 + b2 + c2 = 13 and a3 + b3 + c3 = 27, then what is the value of 1/a+1/b+1/c?

Answer 1

So we have the equations,

$a+b+c=3\\\\a^2+b^2+c^2=13\\\\a^3+b^3+c^3=27$

Consider the first equation.

$a+b+c=3$

We take square of both the sides.

$(a+b+c)^2=3^2\\\\a^2+b^2+c^2+2(ab+bc+ac)=9$

But we have $a^2+b^2+c^2=13.$ Then,

$13+2(ab+bc+ac)=9\\\\ab+bc+ac=-2$

And, we may recall the identity,

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac))$

Then, taking the values for the respective ones,

$27-3abc=3(13-(-2))\\\\27-3abc=45\\\\abc=-6$

Now,

$\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}=\dfrac {ab+bc+ac}{abc}\\\\\\\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}=\dfrac {-2}{-6}\\\\\\\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}=\mathbf {\dfrac {1}{3}}$

Hence $\mathbf {\dfrac {1}{3}}$ is the answer.