If one integer is greater than another integer by 3, and the difference of their cubes is 117, what could be their sum?

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Solution

The correct option is B 7
Use plugging in

$a s (x + 3)^{3} - x^{3} = 117$

$x + 3 \geq 5 a s (x + 3)^{3} \geq 117$

put x = 2
$\Rightarrow 5^{3} - 2^{2} = 125 - 8 = 117$
$\Rightarrow x = 2, x + 3 = 5$
So, sum of both numbers = 7

Alternative Method
$(x + 3)^{3} - x^{3} = 117$
$x^{3} + (3)^{3} + 3 (x) (3) (x + 3) - x^{3} = 117$
$x^{3} + 27 + 27 x + 9 x^{2} - x^{3} = 117$
$\Rightarrow x^{2} + 3 x - 10 = 0$
$\Rightarrow (x + 5) (x - 2) = 0$
x = 2, -5
$\Rightarrow s o e i t h e r 2, 5, o r - 5, - 2$
Thus sum = 7 OR = -7

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