The difference between two positive numbers is $4$ and difference between their cubes is $316$ . Find their product.

21

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Solution

The correct option is C $21$
Let,the two number be $x$ and $y$
Therefore,we get
$x - y = 4$
$x^{3} - y^{3} = 316$
$x - y = 4$
Squaring both sides,
$=> (x - y)^{2} = 4^{2}$
$=> x^{2} + y^{2} - 2 x y = 16$
$=> x^{2} + y^{2} = 16 + 2 x y$
Now,
$x^{3} - y^{3} = 316$
$=> (x - y) (x^{2} + y^{2} + x y) = 316$
$=> 4 (16 + 2 x y + x y) = 316$
$=> 16 + 3 x y = \frac{316}{4}$
$=> 3 x y = 79 - 16$
$=> x y = \frac{63}{3}$
$=> x y = 21$

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