document.write( "Question 273924: Factor completely
\n" ); document.write( "x^6 - 6x - 16\r
\n" ); document.write( "\n" ); document.write( "I factored it to (x^3-8)*(x^3+2) and then to(x-2)*(x^2+2x+4)*(x^3+2) but I don't know how to factor it further. Any help would be appreciated.
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Algebra.Com's Answer #200349 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
As long as the expression is
\n" ); document.write( "\"x%5E6+-+6x%5E3+-+16\"
\n" ); document.write( "and not the
\n" ); document.write( "\"x%5E6+-+6x+-+16\"
\n" ); document.write( "you posted, then you 100% correct. \"%28x-2%29%2A%28x%5E2%2B2x%2B4%29%2A%28x%5E3%2B2%29\" is as far as your expression will factor...

\n" ); document.write( "unless you want to push things a bit and consider 2 to be the perfect cube of \"root%283%2C+2%29\". Then you could use the sum of cubes pattern on the last factor:
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