document.write( "Question 280215: Factor completely sum of cubes
\n" ); document.write( "27x^3+1 \n" ); document.write( "
Algebra.Com's Answer #203652 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "\"27x%5E3%2B1\" Start with the given expression.\r
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\n" ); document.write( "\n" ); document.write( "\"%283x%29%5E3%2B%281%29%5E3\" Rewrite \"27x%5E3\" as \"%283x%29%5E3\". Rewrite \"1\" as \"%281%29%5E3\".\r
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\n" ); document.write( "\n" ); document.write( "\"%283x%2B1%29%28%283x%29%5E2-%283x%29%281%29%2B%281%29%5E2%29\" Now factor by using the sum of cubes formula. Remember the sum of cubes formula is \"A%5E3%2BB%5E3=%28A%2BB%29%28A%5E2-AB%2BB%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%283x%2B1%29%289x%5E2-3x%2B1%29\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "So \"27x%5E3%2B1\" factors to \"%283x%2B1%29%289x%5E2-3x%2B1%29\".\r
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\n" ); document.write( "\n" ); document.write( "In other words, \"27x%5E3%2B1=%283x%2B1%29%289x%5E2-3x%2B1%29\"
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