Question

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 sum of their cubes

• A. $630$
• B. $224$
• C. $539$
• D. $129$

Answer 1

The correct option is C $539$

Given,
$x>y$
Thus, $x-y=5$ and $xy=24$
$x-y=5$
Squaring both sides,
$=>(x-y{)}^2=5^2$
$=>x^2+y^2-2xy=25$
$=>x^2+y^2=25+2\left(24\right)$
$=>x^2+y^2=73$
Now,
$(x+y{)}^2=x^2+y^2+2xy$
$=73+2\left(24\right)$
$=73+48$
$=121$
$=>x+y=\sqrt{121}$
$=\pm 11$
Given,
$x$ and $y$ are positive numbers,
Therefore,
$x+y=11$
Now,
$x^3+y^3=(x+y)(x^2+y^2-xy)$
$=11(73-24)$
$=11\left(49\right)$
$=539$