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Eliminating Multiple Radicals in an Equation

Date: 02/23/2006 at 16:26:22
From: Jim
Subject: eliminating cube roots and higher roots in equations.

Can you simplify an expression that adds cube roots or higher roots,
i.e. eliminate the radicals from the equation?  For example:
  Cuberoot(x) + Cuberoot(y) = 1    or 
  Cuberoot(x) + Fourthroot(y) = 1

It is clear how to proceed with square roots:  move the nonradical 
terms to one side of the equation, square, and repeat; however, with 
cube roots, we end up with cross terms that seem impossible to get 
rid of.

Date: 02/23/2006 at 17:29:01
From: Doctor Vogler
Subject: Re: eliminating cube roots and higher roots in equations.

Hi Jim,

Thanks for writing to Dr. Math.  That's not an easy task, but here is
one way to solve the problem:  First solve for

  cuberoot(y) = 1 - cuberoot(x)


  fourthroot(y) = 1 - cuberoot(x)

and then raise to the 3rd or 4th power to get rid of the root on the
left.  On the right side, if you change cuberoot(x)^3 into x, then you
should have a quadratic polynomial in cuberoot(x).  Solve for it using
the quadratic formula.  This will introduce a square root that we
didn't have before, so we'll deal with that later.  Now raise both
sides to the 3rd power to get rid of the cube root, and then solve for
the square root that arose from the quadratic formula.  Square both
sides, and you are done!

If you have any questions about this or need more help, please write
back and show me what you have been able to do, and I will try to
offer further suggestions.

- Doctor Vogler, The Math Forum 

Date: 02/23/2006 at 21:11:53
From: Jim
Subject: eliminating cube roots and higher roots in equations.

Dr. Vogler:

Thanks for your answer.  Can you tell me if you can extend this idea 
to additional terms, something like: 

  cuberoot(x) + fifthroot(y) + seventhroot(z) = 0

I thought you should solve for the highest power,
  seventhroot(z) = fifthroot(y) - cuberoot(x)

then raise both sides to 7th power, but I end up with cross terms that 
don't seem to be able to simplified.  Then I thought to find the 
lowest common multiple (105) and raise to that power, so for at least 
one term, the radicals disappear.  But as you can see the problem just 
blows up into more cross terms.  Is it impossible?  I'm just looking 
for the kernel of the idea of how this problem would be approached.

Thanks, Jim

Date: 02/24/2006 at 16:40:33
From: Doctor Vogler
Subject: Re: eliminating cube roots and higher roots in equations.

Hi Jim,

Here is a completely different way to solve your problem that works
better in general.  Methods like the one I described last time are
pretty ad-hoc; i.e. they don't generalize well.  Instead, you can use
linear algebra, and this idea will work for polynomial equations in
lots of different roots.

Consider your roots as a, b, c.  That is,

  a = cuberoot(x)
  b = fifthroot(y)
  c = seventhroot(z)

which we would do better to write as

  a^3 = x
  b^5 = y
  c^7 = z

(better, because these are polynomial equations).  Then you want to
convert the equation

  a + b + c = 0

into a polynomial equation in x, y, and z alone.  So the thing to do
is to multiply this equation by every monomial

  a^i * b^j * c^k


  0 <= i < 3
  0 <= j < 5
  0 <= k < 7

and replace every occurrence of a^3 by x, of b^5 by y, of c^7 by z. 
For example, the first few equations would be

  a^2 + ab + ac = 0
  x + a^2*b + a^2*c = 0
  ab + b^2 + bc = 0
  a^2*b + ab^2 + abc = 0

Now if you treat x, y, and z as constants and consider only the
variables a, b, and c, then you will end up with 3*5*7 polynomial
equations in the 3*5*7 monomials

  a^i * b^j * c^k.

In other words, if you consider each monomial as a variable, then you
can write these in matrix form as

  A*v = 0

where A is a square matrix of height and width 3*5*7 which contains
the variables x, y, and z but none of a, b, and c, and v is a column
vector that lists off the 3*5*7 monomials in a, b, and c.  Unless the
only solution to your equations is a = b = c = 0, that means that the
matrix A is not invertible, and therefore

  det(A) = 0.

But the determinant is a polynomial function in the entries of your
matrix, which means that the determinant of A is a polynomial in x, y,
and z.

Conclusion:  If you form the matrix by multiplying your equation by
all of the monomials in the roots, then the determinant of that matrix
gives you a polynomial equation in your variables x, y, and z.

Of course, taking the determinant of a 105-by-105 matrix is no easy
task, so I would recommend having a computer program do the work for you.

- Doctor Vogler, The Math Forum 

Date: 02/26/2006 at 13:48:12
From: Jim
Subject: Thank you (eliminating cube roots and higher roots in equations.)

Thank you, Dr. Vogler.  This is a very interesting approach to the
problem that I don't think I would have found anywhere else.

- Jim
Associated Topics:
College Linear Algebra

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