Plz solve this:x^4-27=0...............?

equation of fourth degree, has 4 roots (solutions) -- 2 real , 2 complex. real roots +(27)^(1/4) , -(27)^(1/4), i.e. plus or minus fourth real root of 27. see other answers, let b = 27^(1/4) , write given equation x^4 - b^4 = 0. recognize difference of 2 squares factors (x² - b²)(x² + b²).

again, first factor difference of 2 squares , factors (x² - b²)= (x+b)(x-b). setting each of these linear factors equal 0 gives (real) answers noted. using complex factorization, second expression factors (x² + b²) = (x + bi)(x - bi) i² = -1. setting each of these factors = 0 gives complex solutions bi , -bi, b fourth root of 27.

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