Prove that

Solution:

To prove: $(2+\sqrt{x})^{4}+(2-\sqrt{x})^{4}=2\left(16+24 x+x^{2}\right)$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

$(a+b)^{4}={ }^{4} C_{0} a^{4}+{ }^{4} C_{1} a^{4-1} b+{ }^{4} C_{2} a^{4-2} b^{2}+{ }^{4} C_{3} a^{4-3} b^{3}+{ }^{4} C_{4} b^{4}$

$\Rightarrow{ }^{4} \mathrm{C}_{0} \mathrm{a}^{4}+{ }^{4} \mathrm{C} 1 \mathrm{a}^{3} \mathrm{~b}+{ }^{4} \mathrm{C} 2 \mathrm{a}^{2} \mathrm{~b}^{2}+{ }^{4} \mathrm{C} 3 \mathrm{a}^{1} \mathrm{~b}^{3}+{ }^{4} \mathrm{C} 4 \mathrm{~b}^{4} \ldots$ (i)

$(a-b)^{4}={ }^{4} C_{0} a^{4}+{ }^{4} C_{1} a^{4-1}(-b)+{ }^{4} C_{2} a^{4-2}(-b)^{2}+{ }^{4} C_{3} a^{4-3}(-b)^{3}+{ }^{4} C_{4}(-b)^{4}$

$\Rightarrow{ }^{4} \mathrm{C}_{0} \mathrm{a}^{4}-{ }^{4} \mathrm{C}_{1} \mathrm{a}^{3} \mathrm{~b}+{ }^{4} \mathrm{C}_{2} \mathrm{a}^{2} \mathrm{~b}^{2}-{ }^{4} \mathrm{C}_{3} \mathrm{ab}^{3}+{ }^{4} \mathrm{C}_{4} \mathrm{~b}^{4} \ldots$ (ii)

Adding (i) and (ii)

$(a+b)^{4}+(a-b)^{7}=\left[{ }^{4} C_{0} a^{4}+{ }^{4} C_{1} a^{3} b+{ }^{4} C_{2} a^{2} b^{2}+{ }^{4} C_{3} a^{1} b^{3}+{ }^{4} C_{4} b^{4}\right]+\left[{ }^{4} C_{0} a^{4}-{ }^{4} C_{1} a^{3} b\right.$

$\left.+{ }^{4} C_{2} a^{2} b^{2}-{ }^{4} C_{3} a b^{3}+{ }^{4} C_{4} b^{4}\right]$

$\Rightarrow 2^{\left[\left(\frac{4 !}{0 !(4-0) !} a^{4}\right)+\left(\frac{4 !}{2 !(4-2) !} a^{2} b^{2}\right)+\left(\frac{4 !}{4 !(4-4) !} b^{4}\right)\right]}$

$\Rightarrow 2\left[(1) a^{4}+(6) a^{2} b^{2}+(1) b^{4}\right]$

$\Rightarrow 2\left[a^{4}+6 a^{2} b^{2}+b^{4}\right]$

Therefore, $(a+b)^{4}+(a-b)^{7}=2\left[a^{4}+6 a^{2} b^{2}+b^{4}\right]$

Now, putting $a=2$ and $b=(\sqrt{x})$ in the above equation.

$(2+\sqrt{x})^{4}+(2-\sqrt{x})^{4}=2\left[(2)^{4}+6(2)^{2}(\sqrt{x})^{2}+(\sqrt{x})^{4}\right]$

$=2\left(16+24 x+x^{2}\right)$

Hence proved.