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# Vietas formula in cubes

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Hello, I posted these questions yesterday, but no one helped me, so i am just posting them agian, sorry.

1. If A, b, and c are the roots of the equation 2x^3-6x^2-15x-3=0, then find (1/(a^2)) + (1/(b^2)) + (1/(c^2))

2. Let a, b, c be the roots of the equation 2x^3+4x^2-5x-3=0. Build a cubic equation which has roots ab, bc, ac

3. The polynomials x^3-ax^2-2x-1 and x^3-3x-2 have a common root. Find all possible values of a.

Feb 15, 2021

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1. You mention the method to be used for the solution, Vieta.

If the roots of the cubic are a, b and c, then

$$\displaystyle (x-a)(x-b)(x-c) = 0, \\ \text{expanding,} \\ x^{3}-x^{2}(a+b+c)+x(ab+bc+ca)-abc=0.$$

So, for the given cubic,

$$\displaystyle a+b+c=3\dots \dots(1) \\ ab+bc+ca=-15/2 \dots\dots(2)\\ abc=3/2\dots \dots (3).$$

$$\displaystyle \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}} =\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}{a^{2}b^{2}c^{2}}.$$

The denominator is easy, just substitute (3).

For the numerator it looks like you have to make use of (2), so,

$$\displaystyle (ab+bc+ca)^{2}=a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2a^{2}bc+2ab^{2}c+2abc^{2}\\= a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c),$$

and now it's just substitution from (1), (2) and (3).

Feb 16, 2021