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Old 2004-04-26, 11:05   #6
jinydu's Avatar
Dec 2003
Hopefully Near M48

175810 Posts

Ok, here's another derivation question:

If you're trying to solve a quadratic equation that (lucky you!) has an integer solution, the expression you obtain using the quadratic formula (after straightforward simplifications) will be an obvious integer. But this doesn't seem to be the case with cubics.

Suppose I have the cubic equation: x^3 + 6x - 20 = 0 (this particular equation was chosen on purpose for relative simplicity).

Now, if I didn't know how to solve cubics, I might try plugging in a few small integer values of x. In this case, I would be lucky, because x = 2 is a solution, and I can easily use polynomial division to find the other two.

But suppose that I instead try to solve the equation using the cubic formula (i.e. Cardano's method). After working through the entire process, I would find that one root of the equation is:

x = cube root(10+sqrt(108)) + cube root(10-sqrt(108)).

My calculator easily verifies that the much more complicated expression above is indeed equal to 2. But my calculator just uses numerical approximations for expressions like sqrt(108). Therefore, it couldn't really be considered a mathematical proof.

Anyway, here's the challenge. Prove that

cube root(10+sqrt(108)) + cube root(10-sqrt(108)) = 2 without knowing a priori that 2 is a solution to the original cubic (and without using trial-and-error with the rational roots theorem).

And in general: The cubic formula tends to give answers that look something like cube root (a+sqrt(b)) + (a-sqrt(b)) + c. In the particular cases where the solution has a simpler form (such as an integer, rational number or expression with only a single radical), how can I get to this simpler form?
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