by Luke53 on Wed Mar 14, 2012 6:46 pm
Hi, I checked the equation and there are no errors, when I solve this eqn. with my graphic calculator (TI89) it gives me two roots 1 and -18, but sadly this thing only gives me the results and not how to get to them.
According to my textbook it should be done like you mentioned by cubing both members. My textbook states that: A^3 - B^3 = 0 is the same as (A - B) * (A^2+AB+B^2) = 0 witch falls apart in A - B = 0 and
A^2 + AB + B^2 = 0 the eqn. A^2 + A B + B^2 has no roots except those that make A and B zero at the same time; but also statisfy A = B. The solutions A^3 = B^3 are also solutions of A = B and vice versa.
The eqn. A^(1/3)+B^(1/3) = c^(1/3) becomes rational if we do the following: Cube both members to get: A + B +3*(A B)^(1/3) * A^(1/3) + B^(1/3) = C or 3(A B C)^(1/3) = C - A - B cubing the last eqn again gives: 27 A B C = (C - A - B)^3.
Very nice so far, but what are A, B and C in the eqn to solve? I suppose A= (x-9)^(1/3); B = (x+26)^(1/3) and C = 5. (but that does not mean I am able solve this yet).
Sorry for errors, I had to translate this text.
Regards.
Luke.