Simplify a rational algebraic expression

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DukiMath

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Hello,
I am having troubles with some of these types of rational algebraic expressions. I cannot simplify this problem. I would be very grateful! Thanks! (P.S not sure to which algebra this is for so sorry)2C53E513-1940-4E60-A550-BADAE215A34F.jpeg
 
Have you tried algebraic long division?

It looks like you might need to start with your paper turned landscape (wide) to accommodate the working out!
 
I would not try long division as it would make me dizzy.

I would try to factor. I would group 2 at a time in the numerator and factor each group. Then I would try to factor that result. Then I would do the same for the denominator.

Please post back your results. Hopefully this method will work. Otherwise try grouping differently!
 
Jomo said:
I would not try long division as it would make me dizzy.

I would try to factor. I would group 2 at a time in the numerator and factor each group. Then I would try to factor that result. Then I would do the same for the denominator.
Click to expand...

Jomo is absolutely right this should always be the first thing to try.

I ought to have said in post#2 that I attempted to factor it myself, and failed. But that might be down to me missing something or making a mistake. I should have pointed this out in the advice that I gave. If you succeed in factoring, it could save a lot of time.
 
I haven't worked it all out, but I was able to factor the denominator, at least, by observing that, for example, if x=y the expression is zero, so (x-y) should be a factor. I just guessed and it worked, but the same sort of thing didn't work (yet) for the numerator.
 
Dr.Peterson said:
I haven't worked it all out, but I was able to factor the denominator, at least, by observing that, for example, if x=y the expression is zero, so (x-y) should be a factor. I just guessed and it worked, but the same sort of thing didn't work (yet) for the numerator.
Click to expand...

I think x=y does work for the numerator too? Very well spotted!

BTW: Here's the expression typed in
(x^3*y - x*y^3 + y^3*z - y*z^3 + z^3*x - z*x^3) / (x^2*y - x*y^2 + y^2*z - y*z^2 + z^2*x - z*x^2)

[math] \frac{x^3y - xy^3 + y^3z - yz^3 + z^3x - zx^3}{x^2y - xy^2 + y^2z - yz^2 + z^2x - zx^2} [/math]
If y=x numerator becomes

[math]x^4 - x^4 + x^3z - xz^3 + z^3x - zx^3[/math]
[math]= (x^4 - x^4) + (x^3z - zx^3) + (z^3x -xz^3)[/math]
[math]= 0[/math]
so (x-y) should be a factor of the numerator too. OP can you use this info?
 
In fact, every factor of the denominator turns out to be a factor of the numerator ... so the final answer depends on either doing some long division, or just making a wild guess that turns out to be right, as I did. (The wild guess was based on symmetry.)
 
Simplify the following:
\(\dfrac{x^3 y - x y^3 + y^3 z - y z^3 + z^3 x - z x^3}{x^2 y - x y^2 + y^2 z - y z^2 + z^2 x - z x^2}\)

A common factor of \(x^3 y - x y^3 - x^3 z + y^3 z + x z^3 - y z^3\) and
\(x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2 \) is \(x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2\),
so \(\dfrac{x^3 y - x y^3 + y^3 z - y z^3 + z^3 x - z x^3}{x^2 y - x y^2 + y^2 z - y z^2 + z^2 x - z x^2} =\)
\(\dfrac{x + y + z) (x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2)}{x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2}\):
\(\dfrac{(x + y + z) (x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2}{x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2}\)

\(\dfrac{(x + y + z) (x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2)}{(x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2)}\)
\( =\dfrac{ x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2}{x^2 y - x y^2 - x^2 z + y^2 z + x z^2 - y z^2}\times(x + y + z) = x + y + z\):
Answer: |
| x + y + z
 
pka got it right, the answer is x+y+z , but i do not get how you got the common factor nor anything after. I would be very grateful if you would to do it step by step. Thanks very much. ( this is how the answer is in the book)received_1962300600569155.jpeg
 
I do not think that pka should give you the answer.

The way it works on this forum (did you read the guidelines?) is that you have to show us your attempt and then we give you hints to go further. Although sometimes we give out answers because the student just needs to see it worked out but I do not feel that this problem is one of them.

I gave you a nice hint in post#3. Did you even try to do what I suggested? You need to be involved in solving your problem.

Please post back, with your work!
 
Jomo said:
I do not think that pka should give you the answer.
The way it works on this forum (did you read the guidelines?) is that you have to show us your attempt and then we give you hints to go further. Although sometimes we give out answers because the student just needs to see it worked out but I do not feel that this problem is one of them.
Click to expand...
When an exercise is as pointless as this one, damn right I will gladly supply answers.
 
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