Here's how to get started...
[tex]\sqrt[3]{3+\sqrt{9 + \frac{125}{27}}} - \sqrt[3]{-3+\sqrt{9 + \frac{125}{27}}}[/tex]
can be rearranged to
[tex]\sqrt[3]{\sqrt{9 + \frac{125}{27}} + 3} - \sqrt[3]{\sqrt{9 + \frac{125}{27}} - 3}[/tex]
and [itex]\LARGE {\sqrt[3]x = x^{1/3}}[/tex]<br />
<br />
so we have<br />
<br />
[tex]\left( \sqrt{9 + \frac{125}{27}} + 3 \right) ^{1/3} - \left( \sqrt{9 + \frac{125}{27}} - 3 \right) ^{1/3}[/tex]<br />
<br />
Doing the math under the radical signs results in<br />
<br />
[tex]\left( \sqrt{\frac{368}{27}} + 3 \right) ^{1/3} - \left( \sqrt{\frac{368}{27}} - 3 \right) ^{1/3}[/tex]<br />
<br />
which is equal to<br />
<br />
[tex]\left( \frac{\sqrt{368}}{\sqrt{27}} + 3 \right) ^{1/3} - \left( \frac{\sqrt{368}}{\sqrt{27}} - 3 \right) ^{1/3} = \left( \frac{\sqrt{368} \sqrt{27}}{\sqrt{27} \sqrt{27}} + 3 \right) ^{1/3} - \left( \frac{\sqrt{368} \sqrt{27}}{\sqrt{27} \sqrt{27}} - 3 \right) ^{1/3} = \left( \frac{\sqrt{9936}}{27} + 3 \right) ^{1/3} - \left( \frac{\sqrt{9936}}{27} - 3 \right) ^{1/3}[/tex]<br />
<br />
[tex]= \left( \frac{\sqrt{9936} + 81}{27} \right) ^{1/3} - \left( \frac{\sqrt{9936} - 81}{27} \right) ^{1/3} = \frac{(\sqrt{9936} + 81)^{1/3} - (\sqrt{9936} - 81)^{1/3}}{3}[/tex]<br />
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So, we need to show that the numerator = 3.<br />
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If you cube the whole term, you'd have to show that the numerator is equal to 27.<br />
<br />
<br />
Cubing the numerator (I'll leave the messy work up to you) results in<br />
<br />
[tex]162 - 45 \left[ (\sqrt{9936} + 81)^{1/3} - (\sqrt{9936} - 81)^{1/3} \right][/tex]<br />
<br />
Notice that the term inside the square brackets is the same term we cubed, so essentially what we have is<br />
<br />
[tex]X^3 = 162 -45X[/tex]<br />
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Solve this cubic equation and you get 3 answers; 1 real and 2 complex.<br />
You should find that the real answer is 3, therefore<br />
<br />
[tex](\sqrt{9936} + 81)^{1/3} - (\sqrt{9936} - 81)^{1/3} = 3[/tex]<br />
<br />
and the original statement is equal to 1.[/itex]