Question

Help me out guys. Got stuck over this algebra question ​

Answer 1

Answer:

Compare LHS and RHS to prove the statement.

Step-by-step explanation:

Given: a + b + c = 0

We have to show that $\frac{1}{1 + x^b + x^{-c}} + \frac{1}{1 + x^c + x^{-a}} + \frac{1}{1 + x^a + x^{-b}} = 1$

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:

$\frac{(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1)}{(1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})}$ = 1

⇒ $(1 + x^c + x^{-a})(1 + x^a + x^{-b}) + (1 + x^b + x^{-c})(x^a + x^{-b} + 1) + (x^b + x^{-c} + 1)(x^c + x^{-a} + 1) = (1 + x^a + x^{-b})(1 + x^b + x^{-c})(1 + x^c + x^{-a})$

We simplify each term and then compare LHS and RHS.

Simplifying the first term:

$(1 + x^c + x^{-a})(1 + x^a + x^{-b})$

= $x^{c + a} + x^{c - b} + x^c + 1 + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 1$

= $x^{c + a} + x^{c - b} + x^{c} + x^{-a - b} + x^{-a} + x^{a} + x^{-b} + 2 \hspace{5mm} \hdots (A)$

Now, we simplify the second term we have:

$(1 + x^b + x^{-c})(x^a + x^{-b} + 1)$

= $x^{a + b} + 1 + x^{b} + x^{a - c} + x^{-b - c} + x^{-c} + x^{a} + x^{-b} + 1$

= $x^{a + b} + x^{b} + x^{a - c} + x^{- c - b} + x^{-c} + x^a + x^{-b} + 2 \hspace{5mm} \hdots (B)$

Again, simplifying $(x^b + x^{-c} + 1)(x^c + x^{-a} + 1)$,

= $x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1$

= $x^{b + c} + x^{b - a} + x^b + x^{-c -a} + x^{-c} + x^c + x^{-a} + 2$(C)

Therefore, LHS = A + B + C

= $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

Similarly. RHS

= $(x^{b + c} + x^{b - a} + x^b + 1 + x^{-c -a} + x^{-c} + x^c + x^{-a} + 1)(x^a + x^{-b} + 1)$

Note that if a + b + c = 0, $\implies x^{a + b + c} = x^0 = 1$

So, RHS = $x^{c + a} + x^{c - b} + 2x^c + x^{-a - b} + 2x^{-a} + 2x^{a} + 2x^{-b} + x^{a + b} + 2x^b + x^{a - c} + x^{-c -b} + x^{b + c} + x^{b - a} + x^{-c -a} + 6$

We see that LHS = RHS.

Therefore, the statement is proved.