# How do I prove that $$\displaystyle\sqrt{{{3}}}{\left\lbrace{5}\sqrt{{{2}}}+{7}\right\rbrace}-\sqrt{{{3}}}{\left\lbrace{5}\sqrt{{{2}}}-{7}\right\rbrace}$$ is

How do I prove that $\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}$ is an integer?

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If we numerically calculate $\sqrt[3]{5\sqrt{2}+7}$ it looks an awful lot like $1+\sqrt{2}$ which is a big hint. Let’s prove that first:
${\left(1+\sqrt{2}\right)}^{3}=1+3\sqrt{2}+3{\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{2}\right)}^{3}=7+5\sqrt{2}$
Similarly,
${\left(-1+\sqrt{2}\right)}^{3}=-1+3\sqrt{2}-3{\left(\sqrt{2}\right)}^{2}+{\left(\sqrt{2}\right)}^{3}=-7+5\sqrt{2}$
Now that we have those cube roots in recognizable form, the rest is easy:
$\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\left(1+\sqrt{2}\right)-\left(-1+\sqrt{2}\right)=2$

###### Not exactly what you’re looking for?
Kendall Wilkinson

$x=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}$
${x}^{3}=14-3\left(\sqrt[3]{50-49}\right)x$
${x}^{3}=14-3x$
${x}^{3}+3x-14=0$
$\left(x-2\right)\left({x}^{2}+2x+7\right)=0$
Since the discriminant of the second bracket is less than 0, we can ignore it, and continue with
$x-2=0$
$x=2$