Question

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:-

• $\sf{a+b+c=3\quad\quad\dots(1)}$

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

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

To find:-

• $\sf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\,?}$

Formulas used:-

• $\sf{(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\quad\quad\dots(i)}$

• $\sf{(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))=a^3+b^3+c^3-3abc\quad\quad\dots(ii)}$

Solution:-

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

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

Square both sides,

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

Expand LHS according to the formula $\sf{(i).}$

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

Perform substitution from $\sf{(2).}$

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

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

Consider the formula $\sf{(ii).}$

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

Perform substitutions 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.

Answer 2

Given:

↬ a+b+c= 3

↬ a²+b²+c²= 13

↬ a³+b³+c³= 27

To Find:

✏ Value of $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

Solution:

We know that,

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

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

Now,

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

$\sf{(3)^{2}=13+2(ab+bc+ca)}$

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

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

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

$\sf{2(ab+bc+ca)=-4}$

$\sf{ab+bc+ca=\dfrac{-\cancel{4}}{\cancel{2}}}$

$\sf{ab+bc+ca=-2}------(1)$

Now,

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

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

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

$\sf{27-3abc=(3)(15)}$

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

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

$\sf{-3abc=18}$

$\sf{abc=\dfrac{\cancel{18}}{-\cancel{3}}}$

$\sf{abc=-6}-----------(2)$

Also,

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

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

From (1) and (2),

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

$\pink{\boxed{\sf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{3}}}}$

Hence, the value of $\bf{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$ is $\green\dfrac{1}{3}$.