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Answers
Answer:
Step-by-step explanation:
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