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Cube and Square Roots

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Date: 1 Mar 1995 19:29:18 -0500
From: Anonymous
Subject: Re: math problems

Dear Dr. Math,
Can you stand some more problems from teachers?  Answers in 24
hours will merit you a bonus!

(1)  What is the sum of the cubes of the roots of 3x^3 + 4x + 2 = 0 ?
(Is there a short way to get this without actually finding
the roots?

(2)  Simplify:  sqrt(3 - sqrt(5))/ (sqrt(2) + sqrt(7 - 3sqrt(5)))

or (3 - (5^.5)^.5)/ (2^.5 + ( 7 - 3*5^.5))^.5

These two are supposed to say the same thing; however, I
have trouble seeing the math symbols all written this way, so
I'm not sure they do.

Thanks,
Janet Ramser
Clarksville, TN
```

```
Date: 1 Mar 1995 23:45:52 -0500
From: Dr. Ken
Subject: Re: math problems

Well, I'll see what I can do.

For the first problem, try this:  every polynomial factors into linear
factors in the complex numbers.  So we can write this polynomial as
3(x-r1)(x-r2)(x-r3).  If you multiply this polynomial out, you'll get
the following polynomial:

-3 r1*r2*r3 + 3(r1*r2 + r1*r3 + r2*r3)x - 3(r1 + r2 + r3)x^2 + 3x^3

So we know that -3(r1*r2*r3) = 2, and so on, in terms of the
coefficients of the original polynomial.

Now, let's say we cube the 3(r1 + r2 + r3) term.  That will give us
r1^3 + r2^3 + r3^3 plus some other garbage.  We want to get rid
of that other garbage, which we can do.  Notice that if we subtract
off 3(r1*r2 + r1*r3 + r2*r3)(r1*r2*r3) and also 3(r1*r2*r3), we'll
get rid of all the garbage!  See if this method pans out for you.

The second one is quite similar to the one that you sent in a little
while ago:  the key will be simplifying the Square root of 7 - 3Sqrt(5)
and the Square root of 3 - Sqrt(5).  Like before, this will involve
systems of equations, being 2ab = 3, a^2 + 5b^2 = 7 and
2cd = 1, c^2 + 5d^2 = 3.  If you solve these equations, you should
unravel the problem.  Let me know if this is too vague or if I just
don't make sense.

-Ken "Dr." Math
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Associated Topics:
High School Basic Algebra

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