Giải pt
c, $\left(x+7\right)\left(x^2-9x+1-\sqrt[3]{20x^2+102x-121}\right)+63x+1=0$
PT $\Leftrightarrow x^{3}-2x^{2}+x+8-(x+7)\sqrt[3]{20x^{2}+102x-121}=0$
$\Leftrightarrow x^{3}-4x^{2}-12x+15+(2x^{2}+13x-7)-(x+7)\sqrt[3]{20x^{2}+102x-121}=0$
$\Leftrightarrow (x^{3}-4x^{2}-12x+15)+(x+7)[2x-1-\sqrt[3]{20x^{2}+102x-121}]=0$
$\Leftrightarrow (x^{3}-4x^{2}-12x+15)+(x+7)\frac{8(x^{3}-4x^{2}-12x+15)}{(2x-1)^{2}+(2x-1)\sqrt[3]{20x^2+102x-121}+(\sqrt[3]{20x^2+102x-121})^2}$
$\Leftrightarrow (x-1)(x^{2}-3x-15)(1+\frac{8(x+7)}{(2x-1)^{2}+(2x-1)\sqrt[3]{20x^2+102x-121}+(\sqrt[3]{20x^2+102x-121})^2})=0$
Dễ dàng CM : $1+\frac{8(x+7)}{(2x-1)^{2}+(2x-1)\sqrt[3]{20x^2+102x-121}+(\sqrt[3]{20x^2+102x-121})^2}>0$ ( Quy đồng tách bình phương)
Vậy nghiệm là $1;\frac{3+\sqrt{69}}{2};\frac{3-\sqrt{69}}{2}$