Question

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

• A. $64$
• B. $37$
• C. $65$
• D. $58$

Answer 1

The correct option is D $58$

Let,the two number be $x$ and $y$, where $x>y$
Therefore, we get
$\Rightarrow x^3-y^3=316$ [Given: Difference between cubes of two numbers is 316]
$\Rightarrow x-y=4$ [Given: Difference between two numbers is 4]
Squaring both sides, we have
$\Rightarrow (x-y{)}^2=4^2$
$\Rightarrow x^2+y^2-2xy=16$
$\Rightarrow x^2+y^2=16+2xy\dots \dots \left(i\right)$
Now,
$x^3-y^3=316$
$\Rightarrow (x-y)(x^2+y^2+xy)=316$
$\Rightarrow 4(16+2xy+xy)=316$ [Using eq.(i)]
$\Rightarrow 16+3xy=\frac{316}{4}$
$\Rightarrow 3xy=79-16$
$\Rightarrow xy=\frac{63}{3}$
$\Rightarrow xy=21$
Putting the value of $xy$ in eq.(i), we get
$\Rightarrow x^2+y^2=16+2\left(21\right)$
$\Rightarrow x^2+y^2=16+42$
$\Rightarrow x^2+y^2=58$