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## Re: [Axiom-developer] Testing if (72*a^3*b^5)^(1/2) is equivalent to 6*a

**From**: |
Martin Rubey |

**Subject**: |
Re: [Axiom-developer] Testing if (72*a^3*b^5)^(1/2) is equivalent to 6*a*b^2*(2*a*b)^(1/2) |

**Date**: |
Mon, 08 Mar 2010 10:18:25 +0100 |

**User-agent**: |
Gnus/5.11 (Gnus v5.11) Emacs/22.3 (gnu/linux) |

Ted Kosan <address@hidden> writes:
>* I have been experimenting with Axiom to see how it compares to other*
>* computer algebra systems.*
>
>* One of the things I tried testing was if Axiom could determine if*
>* (72*a^3*b^5)^(1/2) was equivalent to 6*a*b^2*(2*a*b)^(1/2):*
>
>* (2) -> (72*a^3*b^5)^(1/2) - 6*a*b^2*(2*a*b)^(1/2)*
>
>* +------+*
>* | 3 5 2 +----+*
>* (2) \|72a b - 6a b \|2a b*
>
>
>* When I entered this expression into Wolfram Alpha, it returned 0 as a result.*
>
>* Is Axiom capable of determining if (72*a^3*b^5)^(1/2) is equivalent to*
>* 6*a*b^2*(2*a*b)^(1/2) ?*
It should be. At least FriCAS is:
(1) -> (72*a^3*b^5)^(1/2) - 6*a*b^2*(2*a*b)^(1/2)
+------+
| 3 5 2 +----+
(1) \|72a b - 6a b \|2a b
Type: Expression(Integer)
(2) -> normalize %
(2) 0
Type: Expression(Integer)
Of course, you have to be careful interpreting this result, see
William's answer!
Martin