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 difference of their cubes

• A. $532$
• B. $112$
• C. $560$
• D. $485$

Answer 1

The correct option is D $485$

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,
$x^3-y^3=(x-y)(x^2+xy+y^2)$
$=\left(5\right)[73+24]$
$=5\left(97\right)$
$=485$