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If 3x^2 + x = 1, then find 10x^3 + 27x^6.

 Jan 9, 2021
 #1
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If 3x^2 + x = 1, then find 10x^3 + 27x^6.

 

Hello Guest!

 

\(3x^2 + 1\cdot x - 1=0\)

a          b           c

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-1 \pm \sqrt{1^2-4\cdot 3\cdot (-1)} \over 2\cdot 3}\\ x=-\frac{1}{6}\pm\frac{\sqrt{13}}{6}\)

\(x_1= -\frac{1}{6}+\frac{\sqrt{13}}{6} \\ x_2= -\frac{1}{6}-\frac{\sqrt{13}}{6}\)

 

\(10x^3 + 27x^6\)

\(=10\cdot (-\frac{1}{6}+\frac{\sqrt{13}}{6})^3 + 27\cdot (-\frac{1}{6}+\frac{\sqrt{13}}{6})^6=\color{blue}1\)

 

\(=10\cdot ( -\frac{1}{6}-\frac{\sqrt{13}}{6})^3 + 27\cdot ( -\frac{1}{6}-\frac{\sqrt{13}}{6})^6=\color{blue}1\)

laugh  !

 Jan 10, 2021
edited by asinus  Jan 10, 2021

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