Question 632223
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Your error, I think, was stating that *[tex \LARGE a\ =\ (x\ +\ 5)^3] and *[tex \LARGE b\ =\ 27x^3] when in fact you should have started with *[tex \LARGE a^3\ =\ (x\ +\ 5)^3] and *[tex \LARGE b^3\ =\ 27x^3], which would lead you to *[tex \LARGE a\ =\ (x\ +\ 5)] and *[tex \LARGE b\ =\ 3x]


Then using the pattern *[tex \LARGE (a^3\ -\ b^3)\ =\ (a\ -\ b)\left(a^2\ +\ ab\ +\ b^2\right)] to write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 5)^3\ -\ 27x^3\ =\ \left((x\ +\ 5)\ -\ 3x\right)\left((x\ +\ 5)^2\ +\ 3x(x\ +\ 5)\ +\ 9x^2\right)]


You can take it from here.  You should get a linear binomial factor and a quadratic trinomial on which you will need to use the quadratic formula.  The net result will be one real root and a pair of complex conjugates.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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