Рассмотрим комплексное число:
Найдем его модуль и аргумент:
Запишем число в тригонометрической форме:
Найдем значения кубического корня:
![\sqrt[3]{\rho(\cos \phi+i\sin \phi)} =\left\{\sqrt[3]{\rho}\left(\cos\dfrac{\phi+2\pi k}{3} +i\sin\dfrac{\phi+2\pi k}{3} \right)|k=0;1;2\right\}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%5Crho%28%5Ccos+%5Cphi%2Bi%5Csin+%5Cphi%29%7D+%3D%5Cleft%5C%7B%5Csqrt%5B3%5D%7B%5Crho%7D%5Cleft%28%5Ccos%5Cdfrac%7B%5Cphi%2B2%5Cpi+k%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B%5Cphi%2B2%5Cpi+k%7D%7B3%7D+%5Cright%29%7Ck%3D0%3B1%3B2%5Cright%5C%7D)
![(\sqrt[3]{z})_1=\sqrt[3]{\sqrt{6}}\left(\cos\dfrac{-\frac{\pi }{4}}{3} +i\sin\dfrac{-\frac{\pi }{4}}{3} \right)=\sqrt[6]{6}\left(\cos\left(-\dfrac{\pi }{12}\right)+i\sin\left(-\dfrac{\pi }{12}\right)\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bz%7D%29_1%3D%5Csqrt%5B3%5D%7B%5Csqrt%7B6%7D%7D%5Cleft%28%5Ccos%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%7D%7B3%7D+%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cleft%28-%5Cdfrac%7B%5Cpi+%7D%7B12%7D%5Cright%29%2Bi%5Csin%5Cleft%28-%5Cdfrac%7B%5Cpi+%7D%7B12%7D%5Cright%29%5Cright%29)
![(\sqrt[3]{z})_2=\sqrt[3]{\sqrt{6}}\left(\cos\dfrac{-\frac{\pi }{4}+2\pi }{3} +i\sin\dfrac{-\frac{\pi }{4}+2\pi }{3} \right)=\\=\sqrt[6]{6}\left(\cos\dfrac{\frac{7\pi }{4} }{3} +i\sin\dfrac{\frac{7\pi }{4} }{3} \right)=\sqrt[6]{6}\left(\cos\dfrac{7\pi }{12}+i\sin\dfrac{7\pi }{12}\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bz%7D%29_2%3D%5Csqrt%5B3%5D%7B%5Csqrt%7B6%7D%7D%5Cleft%28%5Ccos%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B2%5Cpi+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B2%5Cpi+%7D%7B3%7D+%5Cright%29%3D%5C%5C%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B%5Cfrac%7B7%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B%5Cfrac%7B7%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B7%5Cpi+%7D%7B12%7D%2Bi%5Csin%5Cdfrac%7B7%5Cpi+%7D%7B12%7D%5Cright%29)
![(\sqrt[3]{z})_3=\sqrt[3]{\sqrt{6}}\left(\cos\dfrac{-\frac{\pi }{4}+4\pi }{3} +i\sin\dfrac{-\frac{\pi }{4}+4\pi }{3} \right)=\\=\sqrt[6]{6}\left(\cos\dfrac{\frac{15\pi }{4} }{3} +i\sin\dfrac{\frac{15\pi }{4} }{3} \right)=\sqrt[6]{6}\left(\cos\dfrac{15\pi }{12}+i\sin\dfrac{15\pi }{12}\right)=\\=\sqrt[6]{6}\left(\cos\dfrac{5\pi }{4}+i\sin\dfrac{5\pi }{3}\right)=\sqrt[6]{6}\left(\cos\left(-\dfrac{3\pi }{4}\right)+i\sin\left(-\dfrac{3\pi }{4}\right)\right)](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bz%7D%29_3%3D%5Csqrt%5B3%5D%7B%5Csqrt%7B6%7D%7D%5Cleft%28%5Ccos%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B4%5Cpi+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B-%5Cfrac%7B%5Cpi+%7D%7B4%7D%2B4%5Cpi+%7D%7B3%7D+%5Cright%29%3D%5C%5C%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B%5Cfrac%7B15%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%2Bi%5Csin%5Cdfrac%7B%5Cfrac%7B15%5Cpi+%7D%7B4%7D+%7D%7B3%7D+%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B15%5Cpi+%7D%7B12%7D%2Bi%5Csin%5Cdfrac%7B15%5Cpi+%7D%7B12%7D%5Cright%29%3D%5C%5C%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cdfrac%7B5%5Cpi+%7D%7B4%7D%2Bi%5Csin%5Cdfrac%7B5%5Cpi+%7D%7B3%7D%5Cright%29%3D%5Csqrt%5B6%5D%7B6%7D%5Cleft%28%5Ccos%5Cleft%28-%5Cdfrac%7B3%5Cpi+%7D%7B4%7D%5Cright%29%2Bi%5Csin%5Cleft%28-%5Cdfrac%7B3%5Cpi+%7D%7B4%7D%5Cright%29%5Cright%29)
При изображении получившийся модуль числа ![\sqrt[6]{6}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B6%7D)
является длиной векторов, а получившиеся аргументы -п/12, 7п/12, -3п/4 - углами, на которые нужно повернуть ось х для ее совмещения с направлением векторов
Решение во вложенииииииииииииииииииииии
При x<0 - -2x
При x=0 - 0
При x>0 - 0
(х+2)·(х-2)+x² +4<span>=х²-4+x² +4=2х²
</span><span> при х= -3. 2х²=2·3·(-3)²=6·9=54
ответ : 54
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