document.write( "Question 1142053: Show that cos 2pi/9 is a root of the equation 8x^3 - 6x + 1=0 \n" ); document.write( "
Algebra.Com's Answer #762717 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!
I think you meant to say \"Show it is NOT a root.\"
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\n" ); document.write( "\n" ); document.write( " \"+pi+\" is a transcendental number - which means neither it, nor any rational multiple of it, is the root of ANY polynomial equation with integer coefficients.
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\n" ); document.write( "It is the very definition of a transcendental number.\r
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\n" ); document.write( "EDIT: Oh, disregard, I just noticed cos in front, you should really write cos(2pi/9)
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\n" ); document.write( "\n" ); document.write( "cos(2pi/9) is approx 0.76604444\r
\n" ); document.write( "\n" ); document.write( "Plugging this into the equation gives 0 out, therefore cos(2pi/9) is a root.\r
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