Question

If $a+b+c=0$, prove that identities $(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b})=9$.

Answer 1

$\Rightarrow (\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c})=\frac{bc(b-a)+ca(c-a)+ab(a-b)}{abc}$

$=\frac{(b-c)(c-a)(a-b)}{abc}$

$\therefore$ The given expression becomes;-

$\Rightarrow$$\frac{1}{abc}\{a(c-a)(a-b)+b(b-c)(a-b)+c(b-c)(c-a)\}$

$\Rightarrow$$a(c-a)(a-b)=a^2(a-b-c)+abc$

$\because b+c=-a$

$\therefore a(c-a)(a-b)=2a^3+abc$

Hence expression;-

$\Rightarrow$$\frac{1}{abc}\{2a^3+2b^3+2c^3+3abc\}$

If $a+b+c=0$, then

$\Rightarrow$$a^3+b^3+c^3=3abc$ (known result)

$\therefore \frac{1}{abc}(3\times 2abc+3abc)$

$=\frac{9abc}{abc}=9$