How do you prove $x^3-3x^2-6x-4$ has a zero of the form $\sqrt[3]a+\sqrt[3]b+\sqrt[3]c$, for distinct positive integers a,b,c
Asked
Active
Viewed 88 times
3 Answers
1
We have formulas to compute the roots of a cubic polynomial.
Use them and you get that it has a real solution $\sqrt[3]{3^2}+\sqrt[3]{3}+\sqrt[3]{1}$.
Now, suppose that we don't know the formulas (But still, we should somehow know how solving cubic goes). We can replace $y=x-1$.
To get $(y+1)^3−3(y+1)^2−6(y+1)−4=y^3-9y+-12$.
We can try to find $y$ in the form $y=a+b$ with $3ab=9$.
Imposing this you get equations for $a^3$ and $b^3$, from where you will get the $a^3=3$ and $b^3=3^2$.
0
To eliminate the quadratic coefficient we substitute $x\leftarrow x+1$ and suspect that one of the cube roots is in fact the integer $c=1$. The substitution gives us $$x^3-9x-12$$ and we look for solutions $\sqrt[3]a+\sqrt[3]b$. Plugging this in gives $$ (a+3\sqrt[3]{ab^2}+3\sqrt[3]{a^2b}+b)-9(\sqrt[3]a+\sqrt[3]b)-12$$ The abundance of threes in this and the desire to make the various roots cancel leads us to the idea of trying things like $a=3$, $b=9$ - and succeed.
Hagen von Eitzen
- 374,180
0
Use a classic method for solving a cubic (not having to remember the formula).
First eliminate the squared term by setting $y=x-1$ to obtain $$y^3-9y-12=0$$
Now we look for solutions of the form $y=v+w$, where $y^3=(v+w)^3=(v^3+w^3)+3vw(v+w)$ which we can rewrite as $$y^3-3vw\cdot y-(v^3+w^3)=0$$
Comparing coefficients we obtain: $3vw=9$ from which we derive $$v^3w^3=27$$ and we also have $$v^3+w^3=12$$ which means that $v^3, w^3$ are the roots of the quadratic $$z^2-12z+27=0=(z-3)(z-9)$$[or use the quadratic formula]
So $v^3=3, w^3=9$ (or the other way about). $y=v+w$, and $x=v+w+1$.
Mark Bennet
- 100,194