MHB Prove Polynomial Roots: a(b) of x^6+x^4+x^3-x^2-1

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The discussion focuses on proving that if \( a \) and \( b \) are roots of the polynomial \( x^4+x^3-1 \), then \( ab \) is a root of the polynomial \( x^6+x^4+x^3-x^2-1 \). Using Vieta's relations, the roots \( a, b, c, \) and \( d \) are analyzed, leading to equations involving sums and products of the roots. By substituting variables for the sums and products, the relationships are manipulated to derive a polynomial equation in terms of \( p \), which represents \( ab \). Ultimately, it is shown that \( p \) satisfies the polynomial \( p^6 + p^4 + p^3 - p^2 - 1 = 0 \), confirming that \( ab \) is indeed a root of the specified polynomial. The proof effectively demonstrates the relationship between the roots of the two polynomials.
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If $$a,\;b$$ are roots of polynomial $$x^4+x^3-1$$, prove that $$a(b)$$ is a root of polynomial $$x^6+x^4+x^3-x^2-1$$.
 
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anemone said:
If $$a,\;b$$ are roots of polynomial $$x^4+x^3-1$$, prove that $$a(b)$$ is a root of polynomial $$x^6+x^4+x^3-x^2-1$$.

you mean ab

Let other 2 roots be c and d

From viete’s relation

We have
a+b+ c+d = - 1 ..1
ab+ac+ad+bc+ bd + cd = 0 ..2
abc+abd+acd+bcd = 0 ..3
abcd = - 1 ..4

now letting s = a+ b, t = c+d, p = ab q = cd we get

s+ t = - 1 or t = -1 –s (1)
p + q + st = 0 …(2)
pt + sq = 0 … (3)
pq = -1 or q = - 1/p

now from (2) p – 1/p –s – s^2 = 0 …(5)
and from (3) –p – ps – s/p =0 => s = - p^2/(p^2+1) …(6)

putting in (5) the value of s we get

p – 1/p +p^2/(p^2+ 1) – p^4/(p^2+1)^2 = 0

or multiplying by (p^2+1) we get

p^6 + p^4 +p^3 – p^2 – 1= 0

so p or ab is a root of it
 
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