Let the roots of the polynomial $P(x) = x^3 − x + 1$ be a, b, c. Construct the polynomial with roots $a^4, b^4, c^4$ and degree 3 I know that $P(x) = x^3 − x + 1$ has a single real root. But I guess it's irrelevant to find it.
What I thought was to change $x$ into ${x}^{\frac{1}{4}}$ so it would satisfy the condition. However, the resulting $Q(x) = x^\frac{3}{4} − x^\frac{1}{4} + 1$ is neither a polynomial nor has a real root.
 A: You are on the right track, if you write $u^{\frac14}=x$ and form an equation in $u$ which can be changed into the form of a polynomial.
So, $$u^{\frac34}-u^{\frac14}+1=0\implies u^{\frac14}(u^{\frac12}-1)=-1$$
$$\implies u^{\frac12}-1=-\frac{1}{u^{\frac14}}$$
Now squaring both sides,
$$u-2u^{\frac12}+1=\frac{1}{u^{\frac12}}$$
$$\implies u+1=\frac{1}{u^{\frac12}}+2u^{\frac12}$$
Squaring again, $$u^2+2u+1=\frac1u+4+4u$$
So the polynomial is $$u^3-2u^2-3u-1=0$$
A: Maybe I was wrong in calculations, one needs to check.
Vieta: $a+b+c=0$, $ab+ac+bc=-1$, $abc=-1$.
Let's do some algebraic manipulations to obtain the same for $a^4$, $b^4$, $c^4$.
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac) \Rightarrow a^2+b^2+c^2=2$$
$$(ab+ac+bc)^2=a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)\Rightarrow a^2b^2+a^2c^2+b^2c^2=1$$
$$(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)\Rightarrow a^4+b^4+c^4=2$$
$$a^2b^2c^2=(abc)^2=1$$
$$(a^2b^2+a^2c^2+b^2c^2)^2=a^4b^4+a^4c^4+b^4c^4+2a^2b^2c^2(a^2+b^2+c^2)\Rightarrow a^4b^4+a^4c^4+b^4c^4=-3$$
It's ok, because there are complex roots in $a,b,c$.
$$a^4b^4c^4=(abc)^4=1$$
Reverse Vieta: $x^3-2x^2+3x-1$ is polynomial with roots $a^4$, $b^4$, $c^4$.
