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Prove that (x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1) = xyz. - Mathematics

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Question

Prove that `(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1) = xyz`.

Theorem
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Solution

Given:

Variables x, y, z are all non-zero and expression to analyze:

`(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1)`

To Prove:

`(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1) = xyz`

Proof:

Step 1: Write the denominator with clearer terms.

`(xy)^-1 = 1/(xy)`

`(yz)^-1 = 1/(yz)`

`(zx)^-1 = 1/(zx)`

So the denominator becomes:

`1/(xy) + 1/(yz) + 1/(zx)`

Step 2: Find the common denominator of the denominator terms.

Common denominator is (xyz). 

Express each term:

`1/(xy) = z/(xyz)`

`1/(yz) = x/(xyz)`

`1/(zx) = y/(xyz)`

So, `1/(xy) + 1/(yz) + 1/(zx) = (z + x + y)/(xyz)`.

Step 3: Substitute the denominator back into the main expression:

`(x + y + z)/((z + x + y)/(xyz)) = (x + y + z) xx (xyz)/(z + x + y)`

Step 4: Recognize that the numerator and denominator inside the fraction are the same addition is commutative, so (x + y + z = z + x + y)

`xyz xx (x + y + z)/(x + y + z) = xyz`

Thus, `(x + y + z)/((xy)^-1 + (yz)^-1 + (zx)^-1) = xyz` is proved.

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Chapter 6: Indices/Exponents - Exercise 6A [Page 129]

APPEARS IN

Nootan Mathematics [English] Class 9 ICSE
Chapter 6 Indices/Exponents
Exercise 6A | Q 12. | Page 129
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