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Question: Simplifying square root term

Question:Simplifying square root term

sunit 105 Maple
simplify square
May 09 2015
1

Hi,

I m trying to simplify an expression involving square roots in Maple. But instead of giving the expected output it is not simplifying it. Please find the attached maple file for reference. I calculated manually and the result is 1. Please help me out for this. Thanx in advance.

Regards

Sunit

> 

restart
> 

temp := (1/2)*(a*r*t-b*p)*sqrt(p+v*sqrt(a*r))*sqrt(b)/(sqrt(a*b*r*t)*sqrt(a*r*t*(p+v*sqrt(a*r))))+(1/2)*(a*r*v*t-v*b*p+p*t*sqrt(a*r)-b*v^2*sqrt(a*r))*sqrt(b)/(sqrt(a*b*r*t*(p+v*sqrt(a*r)))*sqrt(t*(p+v*sqrt(a*r))))+(1/2)*b*(p+v*sqrt(a*r))/(a*r*t)

(1/2)*(a*r*t-b*p)*(p+v*(a*r)^(1/2))^(1/2)*b^(1/2)/((a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2))+(1/2)*(a*r*v*t-v*b*p+p*t*(a*r)^(1/2)-b*v^2*(a*r)^(1/2))*b^(1/2)/((a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2))+(1/2)*b*(p+v*(a*r)^(1/2))/(a*r*t)

 (1)
> 

simplify(temp)

-(1/2)*(-(p+v*(a*r)^(1/2))^(1/2)*b^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a^2*r^2*t^2+(p+v*(a*r)^(1/2))^(1/2)*b^(3/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*p-b^(1/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a^2*r^2*t^2*v+b^(3/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*v*p-b^(1/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t^2*p*(a*r)^(1/2)+b^(3/2)*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t*v^2*(a*r)^(1/2)-b*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*p-b*(a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*v*(a*r)^(1/2))/((a*b*r*t)^(1/2)*(a*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(a*b*r*t*(p+v*(a*r)^(1/2)))^(1/2)*(t*(p+v*(a*r)^(1/2)))^(1/2)*a*r*t)

 (2)
> 

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