Determine all possible real values of x, that satisfy this equation:
(x2+x)3 (x-3)3
------- - ------- =1
(x2+3)3 (x2+3)3
Denominators are the same
x^6 + 3x^5 + 3x^4 + x^3 - x^3 + 9x^2 - 27x + 27 = x^6 + 9x^4 + 27x^2 + 27
3x^5 - 6x^4 - 18x^2 - 27x = 0
x^5 - 2x^4 - 6x^2 - 9x = 0
From summing coefficients of even vs odd powers, we see that x = -1 is a solution
x^5 + x^4 - 3x^4 - 3x^3 + 3x^3 + 3x^2 - 9x^2 - 9x = 0
(x)(x+1)(x^3 - 3x^2 + 3x - 9) = 0 x=3 also works
(x)(x+1)(x-3)(x^2 + 3) = 0
Real solutions: {-1, 0, 3} which all check and this was the requested information.
Complex solutions: {√3i, -√3i} (but I did not check to see if these solve the original equation)
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Posted by Larry
on 2024-08-29 09:14:51 |
Solution