Question

prove that (x^(a-b))^(a+b) . (x^(b-c))^(b+c) . (x^(c-a))^(c+a) = 1

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

LHS :

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

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

=$x^{0}$

= 1 = RHS

Therefore, LHS = RHS [hence, proved].

We first write the given data provided to us in the form of to the power of x

and then we apply law of indices which has a property that when the base will be same and there will be powers given then base will remain constant and the power will be added to each other.

So, terms of a + terms of b + terms of c gets added with each other and then mutually cancelled. (Since + - will give -).

So, the answer was : x gets a power of zero.

and we know that anything to the power zero is always one

Therefore, x^0 = 1

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