Question

(x^a-b)^a+b (x^b-c)^b+c (x^c-a)^c+a = 1​

Answer 1

Step-by-step explanation:

Hello!

Here we know that it is (x^a-b)^a+b (x^b-c)^b+c (x^c-a)^c+a = 1

So

Since (a^m)^n = a^mXn

That can be written as x^(a-b)(a+b) Do it for every term.

And also here we have a identity a²-b²=(a+b)(a-b)

So we can write as x^(a²-b²) x^ (b²-c²) x^(c²-a²)=1

And we know that bases r same power should be added.

a²-b²+b²-c²+c²-a²=0

So x⁰=1

That is true as a⁰=1

Hope it helps !

Thank you! (✷‿✷)