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# Proof Problem #2

If $$a, b$$ are the roots of $$x^4 + x^3 - 1 = 0$$ then prove that $$ab$$ is a root of $$x^6 + x^4 + x^3 - x^2 - 1 = 0$$

Note by Dev Sharma
1 year, 3 months ago

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Suppose $$a,b,c,d$$ are the four roots of $$f(x) = x^{4} + x^{3} - 1.$$ Then $$f(x) = (x - a)(x - b)(x - c)(x - d) =$$

$$x^{4} - (a + b + c + d)x^{3} + (ab + ac + ad + bc + bd + cd)x^{2} - (abc + abd + acd + bcd)x + abcd.$$

Comparing like coefficients gives us that $$a + b + c + d = -1, ab + ac + ad + bc + bd + cd = 0,$$

$$abc + abd + acd + bcd = 0$$ and $$abcd = -1.$$

Now look at the function $$g(x) = (x - ab)(x - ac)(x - ad)(x - bc)(x - bd)(x - cd).$$

When we expand this expression, we can use the equations involving $$a,b,c,d$$ above to establish numerical values for the coefficients of $$g(x).$$ For example, the coefficient for $$x^{5}$$ will be $$-(ab + ac + ad + bc + bd + cd) = 0,$$ and the constant term, i.e., the coefficient of $$x^{0}$$, will be $$(abcd)^{3} = -1.$$ Calculating the other coefficients is quite tedious, (and a bit tricky), but in doing so one will find that $$g(x) = x^{6} + x^{4} + x^{3} - x^{2} - 1,$$ and so the product of any two of the roots of the first equation will be a root of the second equation.

Had you already got this far and were having difficulty calculating the coefficients, or is this enough of a hint to help you finish the proof? · 1 year, 3 months ago

Nice suggestion... · 1 year, 3 months ago

@Dev Sharma A and b are not the only two roots of the equation. · 1 year, 3 months ago

a and b are two roots. But it has other two root also · 1 year, 3 months ago