20040427, 16:31  #23  
"William"
May 2003
New Haven
100100111000_{2} Posts 
Quote:
(a+b*sqrt(3))^3 = 10+6*sqrt(3) the left expands as a^3 + 3a^2b*sqrt(3) + 3ab^2*3+b^3*sqrt(3), so we need a^3+9*a*b^2 = 10 3a^2b+3b^3=6 a*(a^2+9*b^2)=10 b*(a^2+b^2)=2 The second equation b is a factor of 2, so it's +/1 or +/2. Two can't work because the secod term is then 4 or more, so the the only solutions are b=1 a=+/1. Moving to the first equation, a=1 and b=1 is the only solution, and it then allows you to simplify the cube root. Applying to 54+sqrt(2700), you get a(a^2+9b^2)=54 b(a^2+b^2)=10 a is negative from the first equation, b is positive from the second equation. From the second equation, b divides 10 and b^3<10, so b=1 or 2. Both cases have solutions: a=3 or a=1. Checking back to the first equation, only a=3 b=1 works, so the simplification comes from (3+sqrt(3))^3 = 54+sqrt(2700) Note that this works because there were integer solutions to a and b  rational solutions would also simplify the answer but would not have been found. Also the trial and error solution is similar to but less effective than trying all the rational possibilities in the original equation. 

20040427, 16:32  #24  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2^{8}·41 Posts 
Quote:
I really must stop trying to factor small integers in my head. 108=2*2*3*3*3. Doesn't change my argument but the correct factorization removes x=8 from candidates to be considered. Paul 

20040428, 01:05  #25  
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Quote:


20040503, 09:51  #26 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
Sorry if I'm Being Annoying, But
Looks like I've run into another problem like this:
cube root (10+sqrt(108)) = 1 + sqrt(3) cube root ((25/432)+(5sqrt(15)/144)i) = ? This time, I'm not working with "nice integers"... 
20040503, 12:25  #27 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
Never mind...
cube root ((25/432)+(5sqrt(15)/144)i) = (5/12) + (sqrt[15]/12)i 
20040504, 17:21  #28  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Quote:
QUOTE=jinydu]Ok, I'll try to frame my question more clearly this time. Goal: Find a solution of x^3 + 6x  20 = 0 and express it in the simplest possible form. Condition: Not allowed to use trialanderror guessing of rational roots or foreknowledge of the solution. Hint: Applying Cardano's method gives: x = cube root(10+sqrt(108)) + cube root(10sqrt(108)), but this may or may not be the simplest possible way of expressing this solution.[/QUOTE] I will solve the questions in 2 parts. 1) Further simplication of the cube rt. obtained from Cardano's method. 2) Solve the eq. x^5 +6x 20 = 0 This will be done in two posts one for each point. 1) Take the first term in brackets without the cube rt. viz. (10 +sq.rt.108) Now 10+ sq.rt. 108 = 10 + sq.rt (2*2*3*3*3) = 10 + 6sq.rt.3 This can be further be simplified to 1 + sq.rt 3 after taking the cube rt. of 10 + 6 sq rt 3. The second term similarly simplifies to 1  sq.rt. 3. Now adding both terms we get (1 + sq.rt. 3) + (1 sq.rt 3). This gives 2 which is the rational root Please try this out after you have gone thru the general method as given in the foll. worked example. I leave your problem as homework to do. Ques: Find the cube rt. of 72  32 sq.rt.5 (and simplify) Ans: (If) cube rt. of 72  32 sq.rt. 5 = x  sq.rt.y ( why not rt.x  rt.y ?) ( 1 ) Note: I am dispensing of the symbol sq.rt. and calling it just rt. Then , cube rt of 72 + rt.32 = x + rt.y (Theorem) ( 2) By multiplication of (1 ) and (2 )we get cube rt. (5184  (1024 * 5 ) = x^2  y cube rt. 64 = 4 = x^2  y Hence y = x^2  4 (3) Now by removing cube rt. 72  32 rt. 5 = ( x + rt.y ) ^ 3 = ( x^3 + 3xy ) + irrational terms (4) Equating rational terms on both sides we get 72 = x^3 + 3xy From (3) and ( 4) 72 = x^3 +3x ( x^2  4 )) Therefore 72 = x^3 + 3x^3  12x 72 = 4x^3 12x Hence dividing by 4 18 = x^3  3x Therefore x^3  3x  18 = 0 This is a cubic equation in itself which can be solved by Cardano's formula. To cut short here allow yourself to solve by trial which is permissible as in factoring. We find that x = 3 solves the eqn. Hence from (3) y = 5. Therefore the cube root of 72  32 rt 5 = 3  rt 5 or 72 = ( 3  rt. 5) ^ 3 You can check it out by actual multiplying Q.E.D. Mally 

20040531, 10:40  #29 
May 2004
2^{2}·79 Posts 
g.s.
Not thought about it yet.regards
Devaraj It had been worked out geometrically as 1.618033989.... and called phi. Phi was used in building the Great Pyramid of Giza about 3070 B.C. They referred to it as the 'sacred ratio' In the Fibonacci series the ratio of successive terms Fn+1/Fn tends to phi as the series progresses. In dividing a line x+y In parts x,y such that (x+y)/x=x/y Then x/y= (1+sqr.rt.5)/2=1.618033989..... There is a novel way that the ratio can be expressed trigonometrically using the well known constants 'e' and 'i'=sqr.rt (1) The Golden ratio can be shown as 2*cos(log ((i^2))/5*i)) Can anyone show that this is equivalent to phi the golden ratio? Mally.[/QUOTE] 
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