Find all Real Solutions

We find all real solutions x_{1}, ... x_{10} to the equation sqrt{x_{1}^{2} + ... + x_{10}^{2}} = cbrt{x_{1}^{3} + ... + x_{10}^{3}}. The solution involves some satisfying algebra.

00:00 First steps
00:44 Simultaneous equations
03:53 Positive and negative terms
06:26 Finding solutions

[nathanisbored]

in other words, the original equation has no solutions except for the trivial case where everything disappears down to a monomial and you can just cancel out the radicals

[lf]

the final description of the structure of the solutions could be made slightly more concise using the kronecker delta, i.e. x_i = a d(i, j) = a if i == j, 0 else where j is the index of the non-zero term in the 10-tuple and i is free
excellent video

[samueldeandrade]

It would be more elegant to make the difference of equations B - A, getting (a-x_i) in the terms, and simply noticing that such binomials must be greater than zero and so each term (x_i)²(a-x_i) must be zero.

[Happy_Abe]

We could generalize this to any number of terms right?
Nothing here was specific to 10

[GeoffryGifari]

So the solution works for any nonnegative real number a?

[holyshit]

Could you record video how to calculate the sum
sum(binomial(n, 2k)*binomial(k, m), k=m..floor(n/2))
I have the correct result but guy who gave it to me doesn't want to share with me how he has got it
Result of sum proposed by me is s_{n, m} = (binomial(n-m, n-2m) + binomial(n-m-1, n-2m))*2^{n-2m-1}
but i don't know how to get this result

[tomctutor]

Good in the Reals, now how about complex solutions!

[worldnotworld]

Don't understand the a>=0 restriction at the beginning. Square roots have two values, in this case one positive and one negative, no?