Корень(х+3)=кубический корень(4х+12). найти х

$\sqrt{x + 3} = \sqrt[3]{4x + 12} \\ \\ \sqrt[6]{ {(x + 3)}^{3} } = \sqrt[6]{ {(4x + 12)}^{2} } \\ \\ {(x + 3)}^{3} = {(4x + 12)}^{2} \\ {(x + 3)}^{3} = {( \: \: 4(x + 3) \: \: )}^{2} \\ {(x + 3)}^{3} - 16 {(x + 3)}^{2} = 0 \\ {(x + 3)}^{2} \times (x + 3 - 16) = 0 \\ {(x + 3)}^{2} \times (x - 13) = 0 \\ \\$

$1) \: \: \: {(x + 3)}^{2} = 0 \\ \: \: \: \: \: \: \: x + 3 = 0 \\ \: \: \: \: \: \: \: x = - 3$

$2) \: \: \: \: \: x - 13 = 0 \\ \: \: \: \: \: \: \: \: \: x = 13$

ОТВЕТ: - 3 ; 13