If $a, b, c$ and $d$ are natural numbers such that $a^{3} = b^{6}, c^{3} = d^{4},$ and $d - a = 61,$ then the smallest value of $c - b$ is:

617

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122

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239

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593

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Solution

The correct option is A 617
a,b,c,d are natural numbers.
Also,
$a^{3} = b^{6} = x^{6} \Rightarrow a = x^{2}; b = x; x \in i n t e g e r s$
$c^{3} = d^{4} = y^{12} \Rightarrow c = y^{4}; d = y^{3}; y i n i n t e g e r s .$
Given,
$d - a = 614$
$\Rightarrow y^{3} - x^{12} = 61$
So, y & all ohers
Since, y& x are all integers,we can determine their values
$∴ y = 5, x = 8$
$t h e r e f o r e a = * = 8^{2} = 64$
$c = 5^{4} - 625$
$d = 5^{3} = 125$
$∴ c - b = 625 - 8 = 617$

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