If $$a + b = 8$$ and $$a + a^2b + b + ab^2 = 128$$ then the positive value of $$a^3 + b^3$$ is:

Given, $$a + b = 8$$

$$a + a^2b + b + ab^2 = 128$$

$$=$$>  $$a\left(1+ab\right)+b\left(1+ab\right)=128$$

$$=$$>  $$\left(1+ab\right)\left(a+b\right)=128$$

$$=$$>  $$\left(1+ab\right)8=128$$

$$=$$>  $$ 1+ab =\frac{128}{8}$$

$$=$$>  $$1+ab=16$$

$$=$$>  $$ab=15$$

$$\therefore\ $$ $$a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)$$

$$= 8 (a^2+b^2+2ab-3ab)$$

$$=8 \left(\left(a+b\right)^2-3ab\right)$$

$$=8 \left(8^2-3\left(15\right)\right)$$

$$=8 \left(64-45\right)$$

$$=152$$

Hence, the correct answer is Option D