The coefficient of x^-5 in the binomial expansion of ( x+1 x^ 2 3-x^ 1 3+1 - x-1 x-x^ 1 2)^10 where x≠ 0,1 is

$${[\frac{({x}^{\frac{1}{3}}+1) ({x}^{\frac{2}{3}}-{x}^{\frac{1}{3}}+1)}{({x}^{\frac{2}{3}}-{x}^{\frac{1}{3}}+1)}-\frac{(\sqrt{x}-1) (\sqrt{x}+1)}{\sqrt{x} (\sqrt{x}-1)}]}^{10}$$[(x32​−x31​+1)(x31​+1)(x32​−x31​+1)​−x​(x​−1)(x​−1)(x​+1)​]10

$={({x}^{\frac{1}{3}}+1-1-\frac{1}{{x}^{\frac{1}{2}}})}^{10}$=(x31​+1−1−x21​1​)10 $={({x}^{\frac{1}{3}}-{x}^{\frac{-1}{2}})}^{10}$=(x31​−x2−1​)10

${t}_{r+1}={}^{10}{C}_{r}{({x}^{\frac{1}{3}})}^{10-r}{({x}^{\frac{-1}{2}})}^{r}$tr+1​=10Cr​(x31​)10−r(x2−1​)r${ ( -1 )}^{r}$(−1)r

$${t}_{r+1}={}^{10}{C}_{r}({x}^{\frac{10}{3}-\frac{5r}{6}}){(-1)}^{r}$$tr+1​=10Cr​(x310​−65r​)(−1)r

$$\because \frac{10}{3}-\frac{5r}{6}=-5$$∵310​−65r​=−5

$$\frac{5r}{6}=\frac{50}{6} \Rightarrow r=10$$65r​=650​⇒r=10

Hence, ${}^{10}{C}_{10}{(-1)}^{10}=1$10C10​(−1)10=1