Have Time for Cube Roots?

Algebra Level 4

If $f(x) = \frac { 1 }{ \sqrt [ 3 ]{ { x }^{ 2 }+2x+1 } +\sqrt [ 3 ]{ { x }^{ 2 }+x } +\sqrt [ 3 ]{ { x }^{ 2 } } }$

Evaluate $\displaystyle \sum_{n=1}^{2196} f(n)$

1 solution

U Z
Dec 22, 2014

$\boxed{ 1 = 1}$

$(x + 1) - x = 1$

$a^{3} - b^{3} = (a - b)( a^{2} + ab + b^{2})$

$( \sqrt[3]{x + 1} - \sqrt[3]{x})( \sqrt[3]{(x + 1)^{2}} + \sqrt[3]{(x + 1)(x)} + \sqrt[3]{x^{2}}) = 1$

Thus , the $f(x) = \sqrt[3]{x + 1} - \sqrt[3]{x}$

$f(1) + f(2) + ...... + f(2196)$

$\sqrt[3]{2} - \sqrt[3]{1} + \sqrt[3]{3} - \sqrt[3]{2} + ..... + \sqrt[3]{2197} - \sqrt[3]{2196}$

$\sqrt[3]{2197} - \sqrt[3]{1} = 13 -1 = 12$

Did the same except the fact that I didn't square that 1 before my solution. Can I ask what is that? :P

Kartik Sharma - 6 years, 5 months ago

I too did the same $\Large\smile$

Mehul Chaturvedi - 6 years, 5 months ago