document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=559867>Question 559867</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Volumes of Cubes & Dimensions? The combined volumes of the two cubes with integer side lengths are numerically equal to the combined lengths of their edges. What are the dimensions of the cubes? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #363555 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=363555\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let a and b be the side lengths of the cubes. Since a cube has 12 edges, we have\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 12a + 12b = a^3 + b^3\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "The LHS and RHS can both be factored:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 12(a+b) = (a+b)(a^2 - ab + b^2)\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE 12 = a^2 - ab + b^2 \Rightarrow a^2 - ab + b^2 - 12 = 0\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "We can solve this as a quadratic in terms of a:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE a = \frac{b \pm \sqrt{(-b)^2 - 4(1)(b^2 - 12)}}{2(1)}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE = \frac{b \pm \sqrt{-3b^2 + 48}}{2}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "We want -3b^2 + 48 to be a perfect square (then we must also check that it satisfies the constraints for a to be an integer, but we'll do that later). Fortunately, we do not have to check many values of b since -3b^2 + 48 becomes negative when b >= 5. It can be checked that b = 2 and b = 4 are the only values that yield perfect squares. The corresponding values for a are a = 4 and a = 2 respectively. These two are equivalent (4,2 and 2,4) so the dimensions of the cubes are 4 and 2.<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );