А8 помогите пожалуйста!

Решите задачу:

$3x^2-7x-2=0\\D=49+4\cdot3\cdot2=49+24=73\\x_1=\frac{7-\sqrt{73}}{6},\;x_2=\frac{7+\sqrt{73}}{6}\\\frac{x_1^2}{x_2}+\frac{x_2^2}{x_1}=\frac{x_1^3+x_2^3}{x_1x_2}\\x_1^3+x_2^3=\left(\frac{7-\sqrt{73}}{6}\right)^3+\left(\frac{7+\sqrt{73}}{6}\right)^3=\frac{(7-\sqrt{73})^3}{216}+\frac{(7+\sqrt{73})^3}{216}=\\=\frac{343-3\cdot49\sqrt{73}+3\cdot7\cdot73-73\sqrt{73}+343+3\cdot49\sqrt{73}+3\cdot7\cdot73+73\sqrt{73}}{216}=\frac{686+3066}{216}=\\=\frac{3752}{216}$

$x_1x_2=\left(\frac{7-\sqrt{73}}6\right)\left(\frac{7+\sqrt{73}}6\right)=\frac{({7-\sqrt{73}})({7+\sqrt{73}})}{36}=\frac{49-73}{36}=-\frac{24}{36}=-\frac23\\\frac{x_1^2}{x_2}+\frac{x_2^2}{x_1}=\frac{x_1^3+x_2^3}{x_1x_2}=\frac{3752}{216}:\left(-\frac23\right)=-\frac{3752}{216}\cdot\frac32=-\frac{1876}{72}=-26\frac4{72}=-26\frac1{18}$