Question

If $x=1$ is the only real root of $x^3+3x-4=0$, then $(2+\sqrt{5}{)}^{1/3}+(2-\sqrt{5}{)}^{1/3}$ is..?

• A. Rational
• B. Irrational
• C. Complex
• D. None of these

Answer 1

Answer: (A)

Rational

Let $\alpha =(2+\sqrt{5}{)}^{\frac{1}{3}}+(2+\sqrt{5}{)}^{\frac{1}{3}}$
So, $(2+\sqrt{5}{)}^{\frac{1}{3}}+(2-\sqrt{5}{)}^{\frac{1}{3}}-\alpha =0$
Let $a=(2+\sqrt{5}{)}^{\frac{1}{3}}$ and $b=(2-\sqrt{5}{)}^{\frac{1}{3}}$ and $c=-\alpha \Rightarrow a+b+c=0$
Now, $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab+bc-ca)$
So, our choice of $a,b,c$; gives us;
$(2+\sqrt{5})+(2-\sqrt{5})+(-\alpha {)}^3-(3\left)\right(2+\sqrt{5}{)}^{\frac{1}{3}}(2-\sqrt{5}{)}^{\frac{1}{3}}(-\alpha )=0$ (since $a+b+c=0$)
Hence, $4-{\alpha }^3-3(4-5{)}^{\frac{1}{3}}(-\alpha )=0$
ie., $4-{\alpha }^3-3\alpha =0$; ie, ${\alpha }^3+3\alpha -4=0$
But $\alpha =1$ is only real root of $x^3+3x-4=0$ (Given)
Hence, $\alpha =1$
$\Rightarrow (2+\sqrt{5}{)}^{\frac{1}{3}}+(2-\sqrt{5}{)}^{\frac{1}{3}}=\alpha =1$
so, $(2+\sqrt{5}{)}^{\frac{1}{3}}+(2-\sqrt{5}{)}^{\frac{1}{3}}$ is rational