What is the Value of x in the Equation $\sqrt{x^3+1648}-\sqrt{4949-x^3}=75$?

[anemone]

Here is this week's POTW:

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Given that $\sqrt{x^3+1648}-\sqrt{4949-x^3}=75$ for $x\in\Bbb{N}$. Find $x$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)
1. kaliprasad
2. lfdahl

Honorable mention goes to cmath123, as he has his approach correct except for making a small mistake in drawing the final conclusion.

You can find the suggested solution below:

Spoiler

Let $a=\sqrt{x^3+1648}$ and $b=\sqrt{4949-x^3}$. So we have $a-b=75$ and $a^2+b^2=6597$. From $(a-b)^2+2ab=a^2+b^2$ and $(a+b)^2=a^2+b^2+2ab$ we get $ab=486$, $a+b=87$ and $a=81$.

Therefore $81=\sqrt{x^3+1648}\implies x=17$.