If x = 7 and y = 2 then, what is the value of $\frac{x^3-y^3}{x^2-y^2}-\frac{3xy}{x+y}$ ?
Answer & explanation
Correct answer: option 3
x = 7 and y = 2
$\frac{x^3-y^3}{x^2-y^2}-\frac{3xy}{x+y}$
= $\frac{7^3-2^3}{7^2-2^2}-\frac{3 × 7 × 2}{7 + 2}$
= \(\frac{343}{45}\) - \(\frac{42}{9}\)
= \(\frac{125}{45}\) = \(\frac{25}{9}\)