Welcome 2016! Part 10

$\large= 2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 4^{\frac{1}{4}}$ $\large= 2^{\frac{1}{2}} \times 3^{\frac{1}{3}} \times 2^{\frac{2}{4}}$ $\large= 2 \times 3^{\frac{1}{3}}$ $\large = 8^{\frac{1}{3}} \times 3^{\frac{1}{3}}$ $\large = 24^{\frac{1}{3}}$

$\sqrt{2}\times\sqrt[3]{3}\times\sqrt[4]{4}\\={2}^{\frac{1}{2}}\times{3}^{\frac{1}{3}}\times{4}^{\frac{1}{4}}\\={2}^{\frac{1}{2}}\times{3}^{\frac{1}{3}}\times{({2}^{2})}^{\frac{1}{4}}\\={2}^{\frac{1}{2}}\times{3}^{\frac{1}{3}}\times{2}^{\frac{1}{2}}\\={2}\times{3}^{\frac{1}{3}}\\=2\sqrt[3]{3}\\=\sqrt[3]{3(8)}\\=\sqrt[3]{24}$

Substituting $\sqrt[a]{b}$ with the values of the solution, we get that $3+24=\boxed{27}$ .

$2^{1/2}$ × $3^{1/3}$ × $4^{1/4}$ = $b^{1/a}$ .
$2^{1/2}$ × $3^{1/3}$ × $2^{1/2}$ = $b^{1/a}$

2 × $3^{1/3}$ = $b^{1/a}$

$8^{1/3}$ × $3^{1/3}$ = $b^{1/a}$

$24^{1/3}$ = $b^{1/a}$

a+b= 3+24=27