If $\int_a^b(x^2 - x)\, dx = 18, \int_a^b x^3\, dx = 0$, then $a + b$ is equal to

Given $\int_a^b x^3 \, dx = 0$

$\Rightarrow \left[ \dfrac{x^4}{4} \right]_a^b = 0$

$\Rightarrow \dfrac{b^4 - a^4}{4} = 0$

$\Rightarrow b^4 = a^4$

$\Rightarrow b = a$ or $b = -a$

If $b = a$, then $\int_a^b (x^2 - x) \, dx = 0$, which contradicts the given condition $\int_a^b (x^2 - x) \, dx = 18$.

Therefore, $b = -a$.

This directly gives $a + b = 0$.

To find the exact values, substituting $b = -a$ into the first integral:

$\int_a^{-a} (x^2 - x) \, dx = 18$

$\Rightarrow -\int_{-a}^a (x^2 - x) \, dx = 18$

Using properties of definite integrals for even and odd functions:

$- \left( 2 \int_0^a x^2 \, dx - 0 \right) = 18$

$\Rightarrow -2 \left[ \dfrac{x^3}{3} \right]_0^a = 18$

$\Rightarrow -\dfrac{2a^3}{3} = 18$

$\Rightarrow a^3 = -27 \Rightarrow a = -3$

Since $b = -a$, we get $b = 3$.

Hence, $a + b = -3 + 3 = 0$.

Answer: $0$