Sum calculation

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Mohssine

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a,b,c real numbers such as a/b + b/c + c/a = 4 and a/c + c/b + b/a = 5
Calcululate a^3/b^3 + b^3/c^3 + c^3/a^3
 
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Mohssine said:
a,b,c real numbers such as a/b + b/c + c/a = 4 and a/c + c/b + b/a = 5
Calcululate a^3/b^3 + b^3/c^3 + c^3/a^3
Click to expand...
Use the fact:

x^3 + y^3 + z^3 - 3xyz = (x + y +z) * (x^2 + y^2 + z^2 - xy - yz - zx)

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
where is the three equations to solve the the system of three variables a,b and c
 
we change the variable x=a/b and so on we find the sum x + y +z, but how about x^2 + y^2 + z^2
 
how did you explode those two data a/b + b/c + c/a = 4 and a/c + c/b + b/a = 5
 
Mohssine said:
how do you do to find the sum x^2 + y^2 + z^2 - xy - yz - zx
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Hello Mohssine. Those terms will simplify (the subtractions first). Write out the equation Subhotosh provided, replacing x,y,z with a/b, b/c and c/a. You will see cancellations in those subtracted terms and the result will match one of the given values. Continue simplifying. Get a common denominator and combine remaining terms.

Please share your work, if you get stuck.

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Mohssine said:
how did you explode those two data a/b + b/c + c/a = 4 and a/c + c/b + b/a = 5
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You will be substituting 4 and 5 later when you recognize those forms. This will happen after you substitute for x,y,z and start simplifying.

a/c + c/b + b/a = 5 this might be equal 3 not 5?
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No, 5 is correct.

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Subhotosh Khan said:
you did not make a mistake
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I meant a possible mistake in my own work. (I'm trying to be charitable.)

Mohssine has claimed that the given value 5 is incorrect, so they must have already finished the exercise. I would like to see their result because I expect that it's not correct.

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Subhotosh Khan said:
Did you calculate xy & yz & zx?
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first l made x=a/b; y=b/c; z=c/a; the data become: x+y+z=4; 1/x +1/y + 1/z =5
the second equation gives us: xy+yz+yx=5xyz
i used after that the eq x^3 + y^3 + z^3 - 3xyz = (x + y +z) * (x^2 + y^2 + z^2 - xy - yz - zx) but didnt find the solution yet
 
Mohssine said:
first l made x=a/b; y=b/c; z=c/a; the data become: x+y+z=4; 1/x +1/y + 1/z =5
the second equation gives us: xy+yz+yx=5xyz
i used after that the eq x^3 + y^3 + z^3 - 3xyz = (x + y +z) * (x^2 + y^2 + z^2 - xy - yz - zx) but didnt find the solution yet
Click to expand...
you ask if i found xy+yz+yx, i did, its 5xyz but i didnt get its value
 
I'm not rewriting the exercise's expressions.

I'm rewriting the equation provided by Subhotosh, by substituting a/b, b/c, c/a for x,y,z.

For example:

-xy - yz - zx = -(a/b)(b/c) - (b/c)(c/a) - (c/a)(a/b)

Cancelling common factors on the right-hand side yields -5. (Well, it does for me after substitution because I accept their given value of 5.)

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