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lemon14 · Oct 12, 2010, 08:38 AM

$a=\sqrt[3]{3}+\sqrt[3]{4}$
$b=\sqrt[3]{5}+\sqrt[3]{2}$

ebaines · Oct 13, 2010, 06:10 AM

One way to approach this is to consider what the function $y = x^{1/3}$ looks like. Draw it out and you'll see that as x gets larger the function "flattens out" with a smaller slope. Consequently for values of x greater than 1 the difference between the cube root of (x+1) and the cube root of x is smaller than the difference between the cube root of x and the cube root of (x-1). Which means that:

$<br /> \sqrt[3] 5 - \sqrt[3] 4 \ < \ \sqrt[3] 3 - \sqrt[3] 2<br />$

Can you take it from here?