Question

Solve: $\frac{x^3-y^3}{x-y}$ at $x=3$ and $y=2$.

• A. $25$
• B. $21$
• C. $19$
• D. $15$

Answer 1

The correct option is C $19$

Let us substitute $x=3$ and $y=2$ in the given fraction $\frac{x^3-y^3}{x-y}$ as shown below:

$\Rightarrow$$\frac{x^3-y^3}{x-y}=\frac{3^3-2^3}{3-2}=\frac{27-8}{1}=19$

Hence, $\frac{x^3-y^3}{x-y}=19$ at $x=3$ and $y=2$.