Answer:
Compare LHS and RHS to prove the statement.
Step-by-step explanation:
Given: a + b + c = 0
We have to show that [tex]$ \frac{1}{1 + x^b + x^{-c}} + \frac{1}{1 + x^c + x^{-a}} + \frac{1}{1 + x^a + x^{-b}} = 1$[/tex]
We take LCM, simplify the terms and compare LHS and RHS. We will see that LHS = RHS and the statement will be proved.
Taking LCM, we get:
[tex]$ \frac{(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1)}{(1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})} $[/tex] = 1
⇒ [tex](1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1) = (1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})[/tex]
We simplify each term and then compare LHS and RHS.
Simplifying the first term:
[tex](1 + x^c + x^{-a})(1 + x^a + x^{-b})[/tex]
= [tex]$ x^{c + a} + x^{c - b} + x^c + 1 + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 1 $[/tex]
= [tex]$ x^{c + a} + x^{c - b} + x^{c} + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 2 \hspace{5mm} \hdots (A)$[/tex]
Now, we simplify the second term we have:
[tex]$ (1 + x^b + x^{-c})(x^a + x^{-b} + 1) $[/tex]
= [tex]$ x^{a + b} + 1 + x^{b} + x^{a - c} + x^{-b - c} + x^{-c} + x^{a} + x^{-b} + 1 $[/tex]
= [tex]$ x^{a + b} + x^{b} + x^{a - c} + x^{- c - b} + x^{-c} + x^a + x^{-b} + 2 \hspace{5mm} \hdots (B) $[/tex]
Again, simplifying [tex](x^b + x^{-c} + 1)(x^c + x^{-a} + 1)[/tex],
= [tex]$x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1 $[/tex]
= [tex]$ x^{b + c} + x^{b - a} + x^b + x^{-c -a} + x^{-c} + x^c + x^{-a} + 2 $[/tex](C)
Therefore, LHS = A + B + C
= [tex]$ x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6 $[/tex]
Similarly. RHS
= [tex]$ (x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1)(x^a + x^{-b} + 1) $[/tex]
Note that if a + b + c = 0, [tex]$ \implies x^{a + b + c} = x^0 = 1 $[/tex]
So, RHS = [tex]x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6[/tex]
We see that LHS = RHS.
Therefore, the statement is proved.
