find the cubic polynomial

Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
Find the cubic polynomial whose roots are
a) x₁^2, x₂^2, x₃^2
b) 1/(x₁), 1/(x₂), 1/(x₃)
Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?

Quote:

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Originally Posted by apple2009

Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3.
Find the cubic polynomial whose roots are
a) x₁^2, x₂^2, x₃^2
b) 1/(x₁), 1/(x₂), 1/(x₃)
Because we are learning resolvent cubic now, so I think that this may relate to it. Can someone tell me how to find them?

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I tell you the technique

Let as a root of the new polynomial we are looking for . The key is to write the original root as the subject . ie.

Then sub. back to the equation given :





Squaring gives :

I leave the expansion to you .

Try to do the second one by this method .

Quote:

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Originally Posted by apple2009

Let x₁, x₂, x₃ be the roots of f(x)=x^3+7x^2-8x+3. Find the cubic polynomial whose roots are x₁^2, x₂^2, x₃^2

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Using Vieta's relations, we see that , img.top {vertical-align:15%;} . So [Math] x_{1}^{2}+x_{2}^{2}+x_{3}^2= (x_{1}+x_{2}+x_{3})^2-2\left(x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}\right) = (-7)^2-2(-8) = 65
" alt="x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2} = -8[/Math], and . So [Math] x_{1}^{2}+x_{2}^{2}+x_{3}^2= (x_{1}+x_{2}+x_{3})^2-2\left(x_{1}x_{2}+x_{1}x_{3}+x_{3}x_{2}\right) = (-7)^2-2(-8) = 65
" /> and and , which gives . The signs alternate, so the required equation is [LaTeX ERROR: Convert failed] , which simplependulum's equality yields as well.