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Can someone solve this equation for a?

  1. Apr 28, 2005 #1
    Can someone solve this equation for a?

    [tex]b^2a+2bc^2=\frac{a^3b^2}{c^2}+2ba^2+c^2a[/tex]
     
  2. jcsd
  3. Apr 28, 2005 #2

    AKG

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    P(x) = (b/c)²x³ + (2b)x² + (c² - b²)x + (-2bc²)

    We want to find a such that P(a) = 0

    Factor this cubic P, and you'll get something in the form P(x) = (x - a)(dx² + ex + f)

    That a is precisely the one you're looking for. How to factor cubics.
     
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