http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes en-us 2026 Maplesoft, A Division of Waterloo Maple Inc. Maplesoft Document System Wed, 10 Jun 2026 18:59:25 GMT Wed, 10 Jun 2026 18:59:25 GMT The latest comments added to the Post, difference of two cubes http://www.mapleprimes.com/images/mapleprimeswhite.jpg http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes?ref=Feed:MaplePrimes:difference of two cubes:Comments#comment69748 <p>What do you mean by &quot;it&quot;?&nbsp; The difference of two cubes factors:</p> <pre> &gt; factor(a^3 - b^3); </pre> <p><maple>(a-b)*(a^2+a*b+b^2)</maple></p> <p>Or is &quot;it&quot; x, which is 1 + a^3 - b^3?&nbsp; As a polynomial, this is irreducible:</p> <pre> &gt; factor(1 + a^3 - b^3); </pre> <p><maple>1+a^3-b^3</maple></p> <p>If a and b are supposed to be integers, then the fact that x-1 = a^3 - b^3 does not particularly help in factoring x, which may happen to be prime (e.g. for a=7, b=3).</p> The latest comments added to the Post, difference of two cubes 69748 Wed, 17 Sep 2008 07:29:50 Z Robert Israel Robert Israel http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes?ref=Feed:MaplePrimes:difference of two cubes:Comments#comment69747 <p>Here are the factors I'm looking for:</p> <p>(x - 1) =&nbsp; (x^(1/3) - 1)&nbsp; (x^(2/3) + x^(1/3) + 1)</p> <p>Alla</p> The latest comments added to the Post, difference of two cubes 69747 Wed, 17 Sep 2008 08:04:14 Z alla alla http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes?ref=Feed:MaplePrimes:difference of two cubes:Comments#comment82843 <pre> subs(a=x^(1/3),(factor@subs)(x=a^3,x-1)); (1/3) (2/3) (1/3) (x - 1) (x + x + 1) </pre> The latest comments added to the Post, difference of two cubes 82843 Wed, 17 Sep 2008 08:52:41 Z jakubi jakubi http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes?ref=Feed:MaplePrimes:difference of two cubes:Comments#comment82842 <p>How do you wish to tell Maple to &quot;regard&quot; x-1 as a difference of cubes?</p> <p>Substituting x=a^3, factoring, and then inverting that substitution on x, is one way. Here is one other way.</p> <pre> &gt; factor(x-y^3,x^(1/3)); 2 (1/3) (2/3) (1/3) (y + y x + x ) (-y + x ) &gt; eval(%,y=1); (1/3) (2/3) (1/3) (1 + x + x ) (-1 + x ) &gt; factor(x-y^5,x^(1/5)); 4 3 (1/5) 2 (2/5) (3/5) (4/5) (1/5) (y + y x + y x + y x + x ) (-y + x ) &gt; eval(%,y=1); (1/5) (2/5) (3/5) (4/5) (1/5) (1 + x + x + x + x ) (-1 + x ) </pre> The latest comments added to the Post, difference of two cubes 82842 Wed, 17 Sep 2008 09:31:25 Z pagan pagan http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes?ref=Feed:MaplePrimes:difference of two cubes:Comments#comment94061 <p>I wonder why 'factor' seems unable to recognize that 1=1^3, bypassing somehow the automatic simplification, as this seems to be the missing the step to make it work, instead of the current behavior:</p> <pre> factor(x-1,x^(1/3)); x - 1 </pre> The latest comments added to the Post, difference of two cubes 94061 Wed, 17 Sep 2008 18:50:46 Z jakubi jakubi http://www.mapleprimes.com/posts/38679-Difference-Of-Two-Cubes?ref=Feed:MaplePrimes:difference of two cubes:Comments#comment69744 <p>Jakubi's code for factoring (x - 1) as the difference of two cubes works fine, but I am having trouble understanding the Maple code.&nbsp; Could someone please give me a natural language translation, symbol by symbol?</p> <p>The code is</p> <pre> subs(a=x^(1/3),(factor@subs)(x=a^3,x-1)); (1/3) (2/3) (1/3) (x - 1) (x + x + 1) </pre> <p>Many thanks.</p> <p>Alla</p> The latest comments added to the Post, difference of two cubes 69744 Wed, 17 Sep 2008 20:56:31 Z alla alla