document.write( "Question 144549This question is from textbook
\n" ); document.write( ": Thank you in advance for help with this equation. x^2/3+x^1/3-2=0 \n" ); document.write( "
Algebra.Com's Answer #105291 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"x%5E%282%2F3%29%2Bx%5E%281%2F3%29-2=0\" Start with the given equation\r
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\n" ); document.write( "\n" ); document.write( "Let \"u=x%5E%281%2F3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"u%5E2%2Bu-2=0\" Plug in \"u=x%5E%281%2F3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28u%2B2%29%28u-1%29=0\" Factor the left side (note: if you need help with factoring, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "Now set each factor equal to zero:\r
\n" ); document.write( "\n" ); document.write( "\"u%2B2=0\" or \"u-1=0\" \r
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\n" ); document.write( "\n" ); document.write( "\"u=-2\" or \"u=1\" Now solve for u in each case\r
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\n" ); document.write( "\n" ); document.write( "Remember, we let \"u=x%5E%281%2F3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x%5E%281%2F3%29=-2\" or \"x%5E%281%2F3%29=1\" Replace \"u\" with \"x%5E%281%2F3%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x=-8\" or \"x=1\" Cube both sides in each case\r
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\n" ); document.write( "\n" ); document.write( "Notice that when you plug in \"x=-8\" you don't get the correct answer. So \"x=-8\" is an extraneous solution. \r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the solution is \"x=1\"
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