Question

If the product of two numbers is 21 and their difference is 4, then the ratio of the sum of their cubes to the difference of their cubes is:

• A. 185 : 165
• B. 165 : 158
• C. 185 : 158
• D. 158 : 145

Answer 1

The correct option is C 185 : 158

Let x and y be the numbers.
Given:$x–y=4,xy=21$
$x^3-y^3=(x-y)[x^2+y^2+xy]$
$x^3-y^3=(x-y)[x^2+y^2+xy-2xy+2xy]$
$=(x–y)\left[\right(x-y{)}^2+3xy]$
$=4(16+63)$
$=4\times 79$
$=316$
$=x–y=4\Rightarrow x^2+y^2–2xy=16$
$\Rightarrow (x^2+y^2+2xy-2xy-2xy=16$
$\Rightarrow (x+y{)}^2–2xy–2xy=16$
$\Rightarrow (x+y{)}^2=16+4\times 21$
$x+y=10$
$\therefore x^3+y^3=(x+y)[x^2+y^2-xy]$
$=(x+y)[x^2+y^2+2xy-xy-2xy]$
$=(x+y)\left[\right(x+y{)}^2-3xy]$
$=\left(10\right)\left[\right(10{)}^2-(3\times 21]$
$=10(100-63)$
$=370$

$\frac{\text{Sum of cubes}}{\text{Difference of cubes}}=\frac{x^3+y^3}{x^3-y^3}=370:316=185:158$

Therefore, the required ratio is $\frac{185}{158}$.