If 143+23+1=a3+b then (a,b)=

$\frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1} = \sqrt[3]{a} + b$

$\sqrt[3]{4} + \sqrt[3]{2} + 1$

$\sqrt[3]{2} - 1$

$\frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1} \times \frac{\sqrt[3]{2} - 1}{{(\sqrt[3]{2})}^{3} - 1^{3}}$

$= \frac{\sqrt[3]{2} - 1}{2 - 1} = \sqrt[3]{2} - 1 = \sqrt[3]{a} = b$

$a = 2, b = - 1$

If $\frac{1}{\sqrt[3]{4} + \sqrt[3]{2} + 1} = \sqrt[3]{a} + b$ then $(a, b) =$