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Old 2004-04-27, 09:16   #9
xilman's Avatar
May 2003
Down not across

2×3×1,699 Posts

Originally Posted by jinydu
Would it have been possible if you didn't know beforehand that the left-hand side equals 2?

That is, if the question was:

Simplify cube root(10+sqrt(108)) + cube root(10-sqrt(108)) as much as possible (if possible).
I'd do it this way, assuming that I was given only that information and
nothing else:

write x = a + b, where a = cbrt(10+sqrt(108)) and b = cbrt(10-sqrt(108).

Then cube both sides, to get x^3 = (a^3 + b^3) +3ab(a+b) = 20 + 3abx.

Now x^3 = 20+3x * cbrt((10+sqrt(108)) * (10 - sqrt(108)))
or x^3 = 20 +3x * cbrt (100-108)
or x^3= 20 -6x where I took the integer cube root of -8.

This now gives me a polynomial with integer coefficients. It is, of course, the original equation.

It's possible to solve this one by inspection to find the root x=2. Indeed,
the equation is (x-2)(x^2 + 2x +10).

Before you all jump on me for cheating, please remember that the problem as proposed specified that the only information to be assumed known was the sum of the two cube roots. Deducing a reducible cubic from that information is entirely allowable in my view.

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