Question

Expand ${\left(\sqrt{x}+\sqrt{2}\right)}^6$using Binomial Theorem

Short answer

The value of${\left(\sqrt{x}+\sqrt{2}\right)}^6=6\sqrt{2}{x}^{5/2}+40\sqrt{2}{x}^{3/2}+x^3+30x^2+60x+24\sqrt{2}\sqrt{x}+8$

Step-by-step solution

Step 1. Defination

Let $x$and $a$be real number forany positive integer$n$, we have${\left(x+a\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\cdots \cdots +\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+\cdots \cdots +\left(\begin{array}{c}n\\ n\end{array}\right)a^n=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)x^{n-j}{a}^j$

Step 2. Explanation

$\Rightarrow {\left(\sqrt{x}+\sqrt{2}\right)}^6=\left(\begin{array}{c}6\\ 0\end{array}\right){\left(\sqrt{2}\right)}^0{\left(\sqrt{x}\right)}^6+\left(\begin{array}{c}6\\ 1\end{array}\right){\left(\sqrt{2}\right)}^1{\left(\sqrt{x}\right)}^5+\left(\begin{array}{c}6\\ 2\end{array}\right){\left(\sqrt{2}\right)}^2{\left(\sqrt{x}\right)}^4+\left(\begin{array}{c}6\\ 3\end{array}\right){\left(\sqrt{2}\right)}^3{\left(\sqrt{x}\right)}^3+\left(\begin{array}{c}6\\ 4\end{array}\right){\left(\sqrt{2}\right)}^4{\left(\sqrt{x}\right)}^2+\left(\begin{array}{c}6\\ 5\end{array}\right){\left(\sqrt{2}\right)}^5{\left(\sqrt{x}\right)}^1+\left(\begin{array}{c}6\\ 6\end{array}\right){\left(\sqrt{2}\right)}^6{\left(\sqrt{x}\right)}^0$

$\Rightarrow {\left(\sqrt{x}+\sqrt{2}\right)}^6=6\sqrt{2}{x}^{5/2}+40\sqrt{2}{x}^{3/2}+x^3+30x^2+60x+24\sqrt{2}\sqrt{x}+8$