Question

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Answer 1

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