Question

If xy=3 yz=6 zx=2 find x+y+z​

Answer 1

Answer:

Let:

A=xyz

Then we must have:

⇒x=A/(yz)=A5

⇒y=A/(xz)=A8

⇒z=A/(xy)=A2

So we have:

⇒A=xyz=A5(A8)(A2)=A380

Now, we know A≠0 , since none of the pairwise products of its factors was zero. So that means that we can safely divide by A.

So we have:

⇒A280=1

⟹A=±80−−√=±45–√

So:

⇒xyz=A=±45–√

Note that some of the answers to this question have discarded the negative answer. This is wrong. The negative answer is perfectly valid. The positive answer is when x, y, and z are all positive, and the negative answer is when they are all negative. In either case, their pairwise products are positive, as required

Step-by-step explanation:

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Answer 2

Answer:

Incomplete question.

Step-by-step explanation:

We have the formula,

(x + y + z)² = x² + y² + z² + 2(xy + xz + yz)

Since we don't have value of (x²+y²+z²),

it's not possible to find the required answer.

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