Question

Let the degree of polynomial ${(\sqrt{x^5-1}+x)}^9-{(\sqrt{x^5-1}-x)}^9$ be ${}^{\prime }{\alpha }^{\prime }$ then $\frac{\alpha -19}{2}$ is___.

Answer 1

The expression ${(\sqrt{x^5-1}+x)}^9-{\left(\right(\sqrt{x^5-1})-x)}^9$ is a polynomial of degree 21
So $\frac{21-19}{2}=1$
Let $\sqrt{x^5-1}=\lambda$
$(\lambda +x{)}^9-(\lambda -x{)}^9={(}^9{C}_0{\lambda }^9{+}^9{C}_1{\lambda }^{8}x{+}^9{C}_2{\lambda }^7{x}^2+....)-{(}^9{C}_0{\lambda }^9{-}^9{C}_1{\lambda }^{8}x+.....)-{(}^9{C}_0{\lambda }^9{-}^9{C}_1{\lambda }^{8}x+...)=2{(}^9{C}_1{\lambda }^{8}x{+}^9{C}_3{\lambda }^6{x}^3+....)2\left(9\right(x^5-1{)}^{4}x\left){+}^9{C}_3\right(x^5-1{)}^3{x}^3....$
$\Rightarrow$ degree of polynomial = 21.