If $a=26$ and $b=22$, then the value of $\frac{a^3-b^3}{a^2-b^2}-\frac{3 a b}{a+b}$ is
Answer & explanation
Correct answer: option 3
If $a=26$
$b=22$
$\frac{a^3-b^3}{a^2-b^2}-\frac{3 a b}{a+b}$ = $\frac{( a - b ) ( a^2 + b^2 + ab )}{(a + b )(a - b )}-\frac{3 a b}{a+b}$
= $\frac{( a^2 + b^2 + ab )}{(a + b )}-\frac{3 a b}{a+b}$
= \(\frac{a^2 + b^2 - 2ab}{(a + b )}\)
= \(\frac{(a - b)^2}{(a + b )}\) = \(\frac{(26 - 22)^2}{(26 + 22 )}\) = \(\frac{(16)}{(48 )}\) = \(\frac{1}{3}\)