document.write( "Question 80737: 8(c3 - c) = c (6c + 1) \r
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\n" ); document.write( "\n" ); document.write( "8(c^3-c) = c (6c+1)
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Algebra.Com's Answer #57905 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"8%28c%5E3-c%29+=+c+%286c%2B1%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"8c%5E3-8c+=+6c%5E2%2Bc\" Distribute\r
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\n" ); document.write( "\n" ); document.write( "\"8c%5E3-6c%5E2-8c-c+=+0\" Get all terms to one side\r
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\n" ); document.write( "\n" ); document.write( "\"8c%5E3-6c%5E2-9c+=+0\" Combine like terms\r
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\n" ); document.write( "\n" ); document.write( "\"c%288c%5E2-6c-9%29+=+0\" Factor out a c\r
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\n" ); document.write( "\n" ); document.write( "Now factor \"8c%5E2-6c-9\"\r
\n" ); document.write( "\n" ); document.write( "note: this solver uses x instead of c
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Error occurred executing solver 'quadratic_factoring' .

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\n" ); document.write( "\n" ); document.write( "So we have\r
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\n" ); document.write( "\n" ); document.write( "\"c%284c%2B3%29%282c-3%29+=+0\"\r
\n" ); document.write( "\n" ); document.write( "which gives us\r
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\n" ); document.write( "\n" ); document.write( "\"c=0\" or \"4c%2B3=0\" or \"2c-3=0\" Set each individual factor equal to zero\r
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\n" ); document.write( "\n" ); document.write( "\"c=0\" or \"4c=-3\" or \"2c=3\" Now solve for c in each case\r
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\n" ); document.write( "\n" ); document.write( "So our answer is\r
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\n" ); document.write( "\n" ); document.write( "\"c=0\" or \"c=-3%2F4\" or \"c=3%2F2\"
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