#### Question

Show that:

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

#### Solution

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

LHS = `(x^(a-b))^(a+b)(x^(b-c))^(b+c)(x^(c-a))^(c+a)`

`=[x^((a-b)(a+b))][x^((b-c)(b+c))][x^((c-a)(c+a))]`

`=x^((a^2-b^2))x^((b^2-c^2))x^((c^2-a^2))`

`=x^(a^2-b^2+b^2-c^2+c^2-a^2)`

`=x^0`

= 1

= RHS

Is there an error in this question or solution?

Advertisement

Advertisement

Show That: `(X^(A-b))^(A+B)(X^(B-c))^(B+C)(X^(C-a))^(C+A)=1` Concept: Laws of Exponents for Real Numbers.

Advertisement