Question 79641
80 BEING AN EVEN NUMBER IS DIVISABLE BY 2 THUS 80/2=40 NOW 40 IS DIVISABLE BY 2 THUS 40/2=20 NEW 20 IS DIVISABLE BY 2 THUS 20/2=10 NOW 10 IS DIVISABLE BY 2 THUS 10/2=5. WE ACN QUIT NOW BECAUSE 5 IS A PRIME NUMBER. SO WE NOW HAVE 
2*2*2*2=16 & THE SQRT16=4.
THAT TAKES CARE OF THE NUMBER 4SQRT5.
NOW WE ATTACK THE X^9=X^8*X^1 & SQRTX^8=X^4 SO WE HAVE X^4SQRTX
NOW FOR THE Y^9 TERM Y^11=Y^10*Y^1 OR SQRT^11=Y^5SQRTY THUS WE HAVE THE FOLLOWING REDUCED SET OF TERMS: 
SQRT[80(Y^9)(Y^11)]=4(X^4)(Y^5)SQRT(5XY)