If 3x^2 + x = 1, then find 10x^3 + 27x^6.
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\(3x^2 + 1\cdot x - 1=0\)
a b c
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(x = {-1 \pm \sqrt{1^2-4\cdot 3\cdot (-1)} \over 2\cdot 3}\\ x=-\frac{1}{6}\pm\frac{\sqrt{13}}{6}\)
\(x_1= -\frac{1}{6}+\frac{\sqrt{13}}{6} \\ x_2= -\frac{1}{6}-\frac{\sqrt{13}}{6}\)
\(10x^3 + 27x^6\)
\(=10\cdot (-\frac{1}{6}+\frac{\sqrt{13}}{6})^3 + 27\cdot (-\frac{1}{6}+\frac{\sqrt{13}}{6})^6=\color{blue}1\)
\(=10\cdot ( -\frac{1}{6}-\frac{\sqrt{13}}{6})^3 + 27\cdot ( -\frac{1}{6}-\frac{\sqrt{13}}{6})^6=\color{blue}1\)
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