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

143+23+1=a3+b43+23+123−1143+23+1×23−1(23)3−13=23−12−1=23−1=a3=ba=2,b=−1

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

143+23+1=a3+b43+23+123−1143+23+1×23−1(23)3−13=23−12−1=23−1=a3=ba=2,b=−1

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

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$(4, 2)$

$(2, - 1)$

$(3, - 2)$

$(2, 1)$

answer is D.

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Detailed Solution

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

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

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

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

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

$a = 2, b = - 1$

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