Question

solve for 'x'

$27 {x}^{3} + 21x + 8 = 0$


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Answer 1

HERE'S YOUR ANSWER

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$27 {x}^{3} + 21x + 8 = 0 \\ \\ = > 3x(9 {x}^{2} + 7) + 8 = 0 \\ \\ = > 3x(9 {x}^{3} + 7) = - 8 \\ \\ so \\ = > 3x = - 8 \\ = > x = \frac{ - 8}{3} \\ \\ and \\ 9 {x}^{2} + 7 = - 8 \\ = > 9 {x}^{2} = - 8 - 7 \\ = > 9{x}^{2} = -15 \\ = > {x}^{2} = \frac{ - 15}{9} \\ = > {x}^{2} = \frac{ - 5}{3} \\ = > x = \sqrt{ \frac{ - 5}{3} } \\ = > x = \frac{ \sqrt{5} i}{√3}$

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Answer 2

Hope it will help u.....