document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=706642>Question 706642</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 5X^(2/3)-4X^(1/3)-1=0 BY ANOTHER METHOD OTHER THAN SUBSTITUTION THX </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #435306 by </B><A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><B>josgarithmetic(39623)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josgarithmetic><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=435306\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Go ahead!  Your variable is then x^(1/3).  You'll find through solution to quadratic formula that x^(1/3) is either 1 or -(1/5).  So what then is x?\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E%281%2F3%29%29%5E3=x ALIGN=MIDDLE  ALT=\"%28x%5E%281%2F3%29%29%5E3=x\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Do that to 1, and you find x=1.<BR>\n" );
document.write( "Do that to -(1/5), and find <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=%28-%281%2F5%29%29%5E3 ALIGN=MIDDLE  ALT=\"x=%28-%281%2F5%29%29%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" >=? </SPAN>\n" );
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