Prove that : ( xa2 / xb2 ) 1 / a +
b. ( xb2 / xc2 ) 1 / b +
c. ( xc2 / xa2 ) 1 / c + a = 1
Answers
Given: The correct terms are:
(x^a^2 / x^b^2)^1/(a+ b) x (x^b^2 / x^c^2)^1/(b+c) × (x^c^2 / x^a^2 )^1/(c+a) = 1
To find: Prove LHS = RHS
Solution:
- Now we have given the terms:
(x^a^2 / x^b^2)^1/(a+ b) × (x^b^2 / x^c^2)^1/(b+c) × (x^c^2 / x^a^2 )^1/(c+a) = 1
- Taking LHS, we have:
(x^a^2 / x^b^2)^1/(a+ b) × (x^b^2 / x^c^2)^1/(b+c) × (x^c^2 / x^a^2 )^1/(c+a)
- We have the formula:
a^m / a^n = a^(m-n)
- Applying this, we get:
(x^(a^2-b^2))^1/(a+ b) × (x^(b^2-c^2))^1/(b+c) × (x^(c^2-a^2) )^1/(c+a)
- Now we have the formula:
(a-b)(a+b) = a^2 - b^2
- Applying this, we get:
( x^((a-b)(a+b)/(a+ b)) ) × (x^(b-c)(b+c)/(b+c)) × (x^(c-a)(c+a)/(c+a) )
- Now cancelling the terms, we get:
x^(a-b) × x^(b-c) × x^(c-a)
- Now using the formula:
a^m x a^n = a^m+n
x^(a-b+b-c+c-a) = x^0
- Anything raised to power 0 is always 1, so:
x^0 = 1 .......................RHS
- Hence proved
Answer:
So we have proved that:
(x^a^2 / x^b^2)^1/(a+ b) x (x^b^2 / x^c^2)^1/(b+c) × (x^c^2 / x^a^2 )^1/(c+a) = 1
Step-by-step explanation:
it would be better to understand in a written paper