Question

Let $x={\left(\sqrt{50}+7\right)}^{1/3}-{\left(\sqrt{50}-7\right)}^{1/3}$. Then,

• A. $x=2$
• B. $x=3$
• C. $x$ is a rational number, but not an integer
• D. $x$ is an irrational number

Answer 1

Answer: (A)

Given,
$x={\left(\sqrt{50}+7\right)}^{1/3}-{\left(\sqrt{50}-7\right)}^{1/3}$
On cubing both sides, we get
$x^3=\left(\sqrt{50}+7\right)-\left(\sqrt{50}-7\right)-3$
${\left(\sqrt{50}+7\right)}^{1/3}{\left(\sqrt{50}-7\right)}^{1/3}$
$\Rightarrow \left[{\left(\sqrt{50}+7\right)}^{1/3}-{\left(\sqrt{50}-7\right)}^{1/3}\right]$
$\Rightarrow x^3=14-3{\left(50-49\right)}^{1/3}\left(x\right)$
$\Rightarrow x^3=14-3x$
$\Rightarrow x^3+3x-14=0$
$\Rightarrow \left(x-2\right)\left(x^2+2x+7\right)=0$
$\Rightarrow x=2$