Question

If $x+y=4$ and $\frac{1}{x}+\frac{1}{y}=4$, then the value of $x^3+y^3$ is

• A. $52$
• B. $64$
• C. $4$
• D. $25$

Answer 1

The correct option is A $52$

Given that

$x+y=4$

Cubing on both sides

$\Rightarrow (x+y{)}^3=4^3$

$\Rightarrow x^3+y^3+3xy(x+y)=64$

$\Rightarrow x^3+y^3+3xy\left(4\right)=64$

$\Rightarrow x^3+y^3+12xy=64$ ...$\left(1\right)$

Now, it is also given that $\frac{1}{x}+\frac{1}{y}=4$

$\Rightarrow \frac{x+y}{xy}=4$

$\Rightarrow xy=\frac{x+y}{4}$

$\Rightarrow xy=\frac{4}{4}=1$

Substituting this value in $\left(1\right)$ we get

$\Rightarrow x^3+y^3+12=64$

$\Rightarrow x^3+y^3=52$