Mathematics - Binomial Theorem Question with Solution | TestHub

$${\left[\frac{1}{\sqrt{5x^3+1}-\sqrt{5x^3-1}}\right]}^8+{\left[\frac{1}{\sqrt{5x^3+1}+\sqrt{5x^3-1}}\right]}^8$$ After rationalise the polynomial we get $$\begin{array}{rl}& {\left[\frac{1}{\sqrt{5x^3+1}-\sqrt{5x^3-1}}\times \frac{\sqrt{5x^3+1}+\sqrt{5x^3-1}}{\sqrt{5x^3+1}+\sqrt{5x^3-1}}\right]}^8\\ & +{\left[\frac{1}{\sqrt{5x^3+1}+\sqrt{5x^3-1}}\times \frac{\sqrt{5x^3+1}-\sqrt{5x^3-1}}{\sqrt{5x^3+1}-\sqrt{5x^3-1}}\right]}^8\end{array}$$ $${\left[\frac{\sqrt{5x^3+1}+\sqrt{5x^3-1}}{\left(5x^3+1\right)-\left(5x^3-1\right)}\right]}^8+{\left[\frac{\sqrt{5x^3+1}-\sqrt{5x^3-1}}{\left(5x^3+1\right)-\left(5x^3-1\right)}\right]}^8$$ $$\begin{array}{rl}& =\frac{1}{2^8}\left[{\left(\sqrt{5x^3+1}+\sqrt{5x^3-1}\right)}^8+{\left(\sqrt{5x^3+1}-\sqrt{5x^3-1}\right)}^8\right]\\ & =\frac{1}{2^8}\left[\begin{array}{l}{{}^8{C}_0{\left(\sqrt{5x^3+1}\right)}^8+{}^8{C}_4{\left(\sqrt{5x^3+1}\right)}^6{\left(\sqrt{5x^3-1}\right)}^2)}^4{\left(\sqrt{5x^3-1}\right)}^4+\\ {}^8{C}_6(\sqrt{5x^3+1}{}^2(\sqrt{{5x^3-)}^1}{}^6+{}^8{C}_8(\sqrt{5x^3-1}{}^8\end{array}\right])\\ & \frac{1}{2^8}\left[\begin{array}{l}{}^8{C}_0{\left(5x^3+1\right)}^4+{}^8{C}_2{\left(5x^3+1\right)}^3\left(5x^3-1\right)+8_{C_4}\\ {\left(5x^3+1\right)}^2{\left(5x^3-1\right)}^2+\\ {}^8{C}_6\left(5x^3+1\right){\left(5x^3-1\right)}^3+8_{C_8}{\left(5x^3-1\right)}^4\end{array}\right]\\ & \end{array}$$ So, the degree of polynomial is 12 , Now, coefficient of $$\begin{array}{rl}& x^{12}=\left[{}^8{\mathrm{C}}_0{5}^4+{}^8{\mathrm{C}}_2{5}^4+{}^8{\mathrm{C}}_4{5}^4+{}^8{\mathrm{C}}_6{5}^4+{}^8{\mathrm{C}}_8{5}^4\right]\\ & =5^4\times \frac{2^8}{2}=5^4\times 2^4\times \frac{2^2}{2}\\ & =10^4\times 2^3=8(10{)}^4\end{array}$$