If, ${\left(2+\sqrt{3}\right)}^{49}+{\left(\sqrt{3}-2\right)}^{49}=a+b\sqrt{3}$

${\left(\sqrt{3}+2\right)}^{49}+{\left(\sqrt{3}-2\right)}^{49}=a+b\sqrt{3}$

Sum of two binomial expansions

$2\left\{{}^{49}{C}_0{\left(\sqrt{3}\right)}^{49}+{}^{49}{C}_2{\left(\sqrt{3}\right)}^{47}{2}^2+{}^{49}{C}_4{\left(\sqrt{3}\right)}^{45}{2}^4+...........\right\}=a+b\sqrt{3}$

$2\left\{{\left(\sqrt{3}\right)}^{49}+{}^{49}{C}_2{\left(\sqrt{3}\right)}^{47}{2}^2+{}^{49}{C}_4{\left(\sqrt{3}\right)}^{45}{2}^4+...........\right\}=a+b\sqrt{3}$$2{\left(\sqrt{3}\right)}^{49}+2 {}^{49}{C}_2{\left(\sqrt{3}\right)}^{47}{2}^2+2 {}^{49}{C}_4{\left(\sqrt{3}\right)}^{45}{2}^4+...........=a+b\sqrt{3}$

$2{\left(\sqrt{3}\right)}^{49}+ {}^{49}{C}_2{\left(\sqrt{3}\right)}^{47}{2}^3+ {}^{49}{C}_4{\left(\sqrt{3}\right)}^{45}{2}^5+...........=a+b\sqrt{3}$

By comparing both sides, we get

$b\ne 0, a=0$