Question

The difference between two positive numbers is $4$ and difference of their cubes is $316$. Find the sum of their squares.

• A. $58$
• B. $52$
• C. $51$
• D. $60$

Answer 1

The correct option is A $58$

Let, the two number be $x$ and $y$

Therefore,we get

$x-y=4$

$x^3-y^3=316$

$x-y=4$

Squaring both sides,

$=>(x-y{)}^2=4^2$

$=>x^2+y^2-2xy=16$

$=>x^2+y^2=16+2xy$

Now, $x^3-y^3=316$

$=>(x-y)(x^2+y^2+xy)=316$

$=>4(16+2xy+xy)=316$

$=>16+3xy=\frac{316}{4}$

$=>3xy=79-16$

$=>xy=\frac{63}{3}$

$=>xy=21$

Putting the value of $xy$ we get,

$x^2+y^2=16+2xy$

$=>x^2+y^2=16+2\left(21\right)$

$=>x^2+y^2=16+42$

$=>x^2+y^2=58$