Question

If $a^3+b^3+c^3=3abc$ and $a+b+c=0$ show that $\frac{(b+c{)}^2}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}=1$

Answer 1

We have

$a^3+b^3+c^3=3abc$ ..... $\left(1\right)$ and $a+b+c=0$ ..... $\left(2\right)$

Have to prove that,

$\frac{(b+c{)}^2}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}=1$

Now from equation 2 we can write

$a+b+c=0$

or, $(a+b)=-c$ ........ $\left(3\right)$

similarly, $(b+c)=-a$ ......... $\left(4\right)$

and $(c+a)=-b$ ......... $\left(5\right)$

Consider the left hand side,

$\frac{(b+c{)}^2}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}$

$=\frac{(-a{)}^2}{3bc}+\frac{(-b{)}^2}{3ac}+\frac{(-c{)}^2}{3ab}$ [placing the values from the equations 3,4 and 5]

$=\frac{(-a{)}^2}{3bc}\left(\frac{a}{a}\right)+\frac{(-b{)}^2}{3ac}\left(\frac{b}{b}\right)+\frac{(-c{)}^2}{3ab}\left(\frac{c}{c}\right)$ [Multiply and divide the fractions with $abc$ respectively]

$=\frac{(a{)}^3}{3abc}+\frac{(b{)}^3}{3abc}+\frac{(c{)}^3}{3abc}$

$=\frac{(a³+b³+c³)}{3abc}$

$=\frac{\left(3abc\right)}{3abc}$ [from the equation 1]

$=1$

$\therefore \frac{(b+c{)}^2}{3bc}+\frac{(c+a{)}^2}{3ac}+\frac{(a+b{)}^2}{3ab}=1$