# Degree of polynomial :$[x + (x^3 – 1)^{1/2}]^5 + [x – (x^3 – 1)^{1/2}]^5$

What is the degree of the polynomial of the following expression ?

$$[x + (x^3 – 1)^{1/2}]^5 + [x – (x^3 – 1)^{1/2}]^5$$

If I am not very wrong the highest power of x is $\frac {15}{2}$ ?! So the degree is floor(15/2) = 7 ?! The answer is 7 but I am not sure of this approach.

-
$(a+b)^5+(a-b)^5=2a^5+20a^3 b^2+10ab^4$ – J. M. Oct 26 '10 at 4:59

Usually, a polynomial is understood to have only non-negative powers of $x$ (i.e. $x^0=1,x^1=x,x^2$ etc.), so a-priori it's not even clear that this is a polynomial. However, if we open up the binomial theorem, we see that the odd powers cancel:

\begin{equation*} (x+y)^n + (x-y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} [y^k + (-y)^{k}]. \end{equation*}

When $k$ is odd, $y^k + (-y)^k$ cancels. In your case, the term corresponding to $k = 2l$ has degree $n-2l+3l = n+l$, and so the maximum is achieved for $l=2$ and $n+l = 7$.

-
Thanks ,for showing me the general formula :) – Tretwick Marain Oct 26 '10 at 5:06

If you expand the 5th powers, the terms that have square roots in them will cancel.

-