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:
\sqrt[3] 5 - \sqrt[3] 4 \ < \ \sqrt[3] 3 - \sqrt[3] 2
Can you take it from here?