Question 242528
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 54x^3\ +\ 2]


Take a 2 out of both terms:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\left(27x^3\ +\ 1\right)]


Then factor the binomial using the sum of two cubes factorization


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a^3\ +\ b^3\ =\ (a\ +\ b)(a^2\ -\ ab\ +\ b^2)]


Hint:  *[tex \LARGE 27x^3\ = (3x)^3] and *[tex \LARGE 1\ =\ 1^3]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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