Algebra , Polynomials

$\begin{aligned} \frac { { \left( b+c \right) }^{ 2 } }{ bc } +\frac { { \left( c+a \right) }^{ 2 } }{ ca } +\frac { { \left( a+b \right) }^{ 2 } }{ ab } &=\frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab } \\ \\ &=\frac { { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 } }{ abc } \\ \\ &=\frac { 3abc }{ abc } = 3 \quad \quad (\color{#3D99F6}{\text{Because if } a + b + c = 0, \text{then } a^3 + b^3 + c^3 = 3abc}) \end{aligned}$

Given

a+b+c=0---->(1)

From (1)

b+c=-a

c+a=-b

a+b=-c

Now

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

$\Rightarrow$ $\frac {(a)^{2}}{bc}$ + $\frac {(b)^{2}}{ca}$ + $\frac {(c)^{2}}{ca}$

$\Rightarrow$ $\frac {a^{4}bc + b^{4}ca + c^{4}ab}{(abc)^{2}}$

$\Rightarrow$ abc( $\frac {a^{3} + b^{3}+ c^{3}}{(abc)^{2}}$ )

$\Rightarrow$ $\frac {a^{3} + b^{3}+ c^{3}}{abc}$ ---->(2)

When a + b + c=0 then $a^{3}$ + $b^{3}$ + $c^{3}$ =3abc

By substituting $a^{3}$ + $b^{3}$ + $c^{3}$ =3abc in (2), we get

$\frac {3abc}{abc}$ =3

Therefore $\frac {(b+c)^{2}}{bc}$ + $\frac {(c+a)^{2}}{ca}$ + $\frac {(a+b)^{2}}{ab}$ when a + b + c = $\boxed{3}$