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anonymous
 5 years ago
Suppose P(x) = x^3  a x^2 + b x  c has three distinct real roots, r1, r2, and r3. What cubic polynonial with 1 as its leading coefficient has r1^4, r2^4, and r3^4 as its roots?
anonymous
 5 years ago
Suppose P(x) = x^3  a x^2 + b x  c has three distinct real roots, r1, r2, and r3. What cubic polynonial with 1 as its leading coefficient has r1^4, r2^4, and r3^4 as its roots?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(x  r1)(x  r2)(x  r3) = x^3  ax^2 + bx  c x^3  (r1 + r2 + r3)x^2 + (r1r2  r3r1  r3r2)x  r1r2r3 = x^3  ax^2 + bx  c thus r1 + r2 + r3 = a [1] r1r2  r3r1  r3r2 = b [2] r1r2r3 = c [3] (x  r1^4)(x  r2^4)(x  r3^4) = x^3  (r1^4 + r2^4 + r3^4)x^2 + (r1^4*r2^4  r3^4*r1^4  r3^4*r2^4)x  r1^4*r2^4*r3^4 We can see from equations [1], [2] and [3], if we raise both sides of [3] to the power of 4: r1^4 * r2^4 * r3^4 = c^4 so last term of the polynomial is c^4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0struggling to see how to express the other terms in terms of a, b and c

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0isn't the answer just P(x)=(xr1^4)(xr2^4)(xr3^4)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The coefficients of the polynomial we're looking for are to be expressed in terms of a, b, and c.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0requires an algebraic trick that i can not see...give me some time :) i get a cubic in r3 that looks hard to solve

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0r1 + r2 + r3 = a r1r2 + r1r3 + r2r3 = b r1r2r3 = c we need to get expressions for similar iidenties as the above replacing r1 by r1^4 etc in terms of a, b an c but that seems really dificult

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to find r1^4 + r2^4 + r3^4 in terms of the three identities and then substitute a, b and c and similarly for the other two identities  thats a daunting task

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0tough one isn't it, jimmy?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm trying to expand r1^4 + r2^4 + r3^4  4r1r2r3 (similar to what you do with a cubic) but its a real headache)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0best of luck!  must go  i have to take the wife shopping.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol  u must be joking!  i'm spending money!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i give up. do you have a solution abtrehearn?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0\[x^3x^2*(a^4+R_a)+x(b^4R_b)c^4\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0lol \[R_a\] means the remaining part of a^4 \[R_b\] means the remaining part of b^4

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0a^4=r1^4+r2^4+r3^4+....+some other stuff there is a minus sign in front of this so i need to add the remaining stuff back

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0so there should be a negative in front of R_a

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0since there is already a negative in front of that parenthesis

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm not sure what R_a represents? can you find the remaining two coefficients in terms of a, b and c?

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0R_a is the ramaining part of a^4 remember a^4=(r1+r2+r3)^4=r1^4+r2^4+r3^4+alot of other gibberish this gibberish is R_a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes, but the gibberish is in terms of r1, r2 and r3

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0(r1+r2+r3)^4=r1^4+r2^4+r3^4+R(x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0One thing that comes in handy is this identity: \[r _{1}^{3} + r _{2}^{3} + r _{3}^{3}  3r _{1}r _{2}r _{3} = (r _{1} + r _{2} + r _{3})(r _{1}^{2} + r _{2}^{2} + r _{3}^{2}  r _{1} r _{2}  r _{1}r _{2}r _{2}r _{3})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[= (r _{1} + r _{2} + r _{3})(r _{1}^{2} + r _{2}^{2} + r _{3}^{2}  r _{1}r _{2}  r _{1}r _{3}  r _{2}r _{3})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[= a(r _{1}^{2} + r _{2}^{2} + r _{3}^{2}  b)\] \[= r _{1}^{3} + r _{2}^{3} + r _{3}^{3}  3c\] Another useful identity is \[\[(r _{1} + r _{2} + r _{3})^{2} = r _{1}^{2} + r _{2}^{2} + r _{3}^{2} + 2(r _{1}r _{2} + r _{1}r _{3} + r _{2}r _{3})\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This identity leads to \[a ^{2} = r _{1}^{2} + r _{2}^{2} + r _{3}^{2} + 2b.\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So a^2  2b can sub in for r1^2 + r2^2 + r3^2 in the cubic identity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If we have x^3 = a x^2  b x + c for x in {r1, r2, r3}, the same is true for x^4 = a x^3  b x^2 + c x = a(ax^2  bx + c) + b x^2 + c x = (a^2 + b) x^2 + (a b + c)
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