Two positive numbers $x$ and $y$ are such that $x > y$ . If the difference of these numbers is $5$ and their product is $24$ , find difference of their cubes

532

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112

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560

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485

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Solution

The correct option is D $485$
Given,
$x > y$
Thus, $x - y = 5$ and $x y = 24$
$x - y = 5$
Squaring both sides,
$=> (x - y)^{2} = 5^{2}$
$=> x^{2} + y^{2} - 2 x y = 25$
$=> x^{2} + y^{2} = 25 + 2 (24)$
$=> x^{2} + y^{2} = 73$
Now,
$x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})$
$= (5) [73 + 24]$
$= 5 (97)$
$= 485$

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