Question

If a+b+c=3 ,a^2+b^2+c^2=13 and a^3+b^3+c^3=27 ,then value of 1/a+1/b+1/c is

Answer 1

Given:-

• $\sf{a+b+c=3\quad\quad\dots(1)}$

• $\sf{a^2+b^2+c^2=13\quad\quad\dots(2)}$

• $\sf{a^3+b^3+c^3=27\quad\quad\dots(3)}$

Consider the equation $\sf{(1).}$

$\displaystyle\longrightarrow\sf{a+b+c=3}$

Square both sides,

$\displaystyle\longrightarrow\sf{(a+b+c)^2=3^2}$

$\displaystyle\longrightarrow\sf{a^2+b^2+c^2+2(ab+bc+ca)=9}$

From $\sf{(2),}$

$\displaystyle\longrightarrow\sf{13+2(ab+bc+ca)=9}$

$\displaystyle\longrightarrow\sf{ab+bc+ca=-2\quad\quad\dots(4)}$

We know the formula given below.

$\displaystyle\longrightarrow\sf{(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))=a^3+b^3+c^3-3abc}$

From $\sf{(1),\ (2),\ (3)}$ and $\sf{(4),}$

$\displaystyle\longrightarrow\sf{3(13-(-2))=27-3abc}$

$\displaystyle\longrightarrow\sf{45=27-3abc}$

$\displaystyle\Longrightarrow\sf{abc=-6\quad\quad\dots(5)}$

Now we have,

$\displaystyle\longrightarrow\sf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{ab+bc+ca}{abc}}$

From $\sf{(4)}$ and $\sf{(5),}$

$\displaystyle\longrightarrow\sf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{-2}{-6}}$

$\displaystyle\longrightarrow\underline{\underline{\sf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}=\bf{\dfrac{1}{3}}}}$

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