Question 36358
1.If α, b, g are the roots of the equation

x^3─7x^2 + x + 5 = 0

Find the equation whose roots are

α^2 + b^2, b^2 + g^2, g^2 + α^2. 

Sol: since α + b+ g= 7,
  αb + bg+ gα = 1 and
 αbg = - 5.
 use α^2+ b^2 + g^2 = (α+b+g)^2 - 2(αb + bg+ gα) = 49 - 2 = 47.
  (α^2 + b^2)(b^2 + g^2)+ (b^2 + g^2)(g^2 + α^2) + (g^2 + α^2)(α^2 + b^2)
 = (47- g^2)(47- α^2) + (47- α^2)(47- b^2) + (47- b^2)(47- g^2)
 = 3*47^2 - 47*2(α^2+ b^2 + g^2) + α^2 g^2 + α^2 b^2 + b^2 g^2
 where α^2 g^2 + α^2 b^2 + b^2 g^2 = (αb + bg+ gα)^2 - 2αbg(α + b+ g)
 = 1 + 10*7 =  71.
 So, (α^2 + b^2)(b^2 + g^2)+ (b^2 + g^2)(g^2 + α^2) + (g^2 + α^2)(α^2 + b^2)
  = 3*47^2 - 2*47^2 + 71 = 2280
 
 And, (α^2 + b^2)(b^2 + g^2)(g^2+ α^2)
 = (47- α^2)(47- g^2) (47- b^2)
= 47^3 - 47(α^2+ b^2 + g^2) + 47(α^2 b^2 + g^2b^2+ α^2g^2)- α^2 b^2 g^2
= 47^3 - 47^2 + 47*71- 25 = 104926

 Hence, the required equation as:
 x^3 - 47x^2 + 2280 x - 104926 = 0.


2.Find all the fifth roots of (2+i).

 Use de Moevie(??) law;
 2+i = r (cos t+ i sin t), where r = sqrt(2^2+1) = sqrt(5)
 and t = ArcTan 1/2
 the five 5th roots are r^(1/5) (cos {{{(2pi k + t)/ 5)}}}  + 
 i sin {{{(2pi k + t)/ 5 }}} }
  for k = 0,1,2,3,4.
 
 or 5^(1/10) (cos {{{t/ 5)}}} +  i sin {{{t/ 5 }}} },
 5^(1/10) (cos {{{(2pi  + t)/ 5)}}}  +  i sin {{{(2pi  + t)/ 5 }}} },
 5^(1/10) (cos {{{(4pi  + t)/ 5)}}}  + i sin {{{(4pi  + t)/ 5 }}} },
 5^(1/10) (cos {{{(6pi  + t)/ 5)}}}  + i sin {{{(6pi  + t)/ 5 }}} } ,
 and
  5^(1/10) (cos {{{(8pi  + t)/ 5)}}}  + i sin {{{(8pi  + t)/ 5 }}} }



 
  
 Kenny