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I had this question:

"Find the cubic equation whose roots are the the squares of that of $x^3 + 2x + 1 = 0$" and I kind of solved it. In that my answer was $x^3 - 4x^2 + 4x + 1$, but it was actually $x^3 + 4x^2 + 4x - 1 = 0$.

I took the general equation of a cubic equation, which was: $x^3 +bx^2/a + cx/a + d/a$.

Through simultaneous equations, I found what $b/a, c/a, d/a$ should equate to for my unknown cubic polynomial. Am I supposed to make $b/a, c/a, d/a$ all positive, then substitute it into the general formula?

Any help would be greatly appreciated, thanks.

mau
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missiledragon
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  • http://en.wikipedia.org/wiki/Vieta's_formulas and http://en.wikipedia.org/wiki/Newton%27s_identities – Ma Ming May 16 '13 at 08:43

2 Answers2

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Let $a$ be a root of $x^3+2x+1=0$ and $b=a^2$ be a root of the required equation

So, $a^3+2a+1=0\implies a\cdot b+2a+1=0\implies a=-\frac1{b+2}$

As $a$ be a root of $x^3+2x+1=0$, put this value of $a$ in $x^3+2x+1=0$

On simplification, I get $(b+2)^3-2(b+2)^2-1=0\iff b^3+4b^2+4b-1=0$

So, the required equation will be $y^3+4y^2+4y-1=0$

lab bhattacharjee
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1

Let $a,b,c$ be roots of $x^3 + 2x + 1 = 0$

Using Vieta's Formulas, $a+b+c=0,ab+bc+ca=2,abc=-1$

So, we need to find the cubic equation whose roots are $a^2,b^2,c^2$

i.e., $$(y-a^2)(y-b^2)(y-c^2)=0$$ $$\iff y^3-(a^2+b^2+c^2)y^2+(a^2b^2+b^2c^2+c^2a^2)y+a^2b^2c^2=0$$

Now, $a^2b^2c^2=(abc)^2=(-1)^2=1$

$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=0-2(2)=-4$

$a^2b^2+b^2c^2+c^2a^2=(ab)^2+(bc)^2+(ca)^2$ $=(ab+bc+ca)^2-2(ab\cdot bc+bc\cdot ca+ca\cdot ab)$ $=(ab+bc+ca)^2-2(a+b+c)abc=2^2-2\cdot0\cdot(-1)=4$

lab bhattacharjee
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