Algebra Precalculus 
$a> 0 < b$  For all the integer value of $a$ and $b$
$X= (a^2 + ab)-(ab^2-b)/(2a^2+b^2 -ab)$
Quantity I: $x $
  Quantity II: $1.5$

(a) Quantity I $\lt$ Quantity II
(b) Quantity I $\gt$ Quantity II
(c) Quantity I $\ge$ Quantity II
(d) Quantity I $=$ Quantity II
(e) No relation 

$(x^a)^c = x^c$
$x^{2b}/x^a = (x^{5a}) * (x^d)*(x^b)$
Quantity I = $b$
Quantity II = $d$

(a) Quantity I $\gt$ Quantity II
(b) Quantity I $\lt$ Quantity II
(c) Quantity I $\ge$ Quantity II
(d) Quantity I $=$ Quantity II
(e) No relation
Generally How to solve this sum and how to approach this Question,please guide me the steps with the answer 

I assume $a = 1$
Then,
$$x^{2b}/x^a = (x^{5a}) * (x^d)*(x^b) \implies x^{2b}/x = (x^{5}) * (x^d)*(x^b) \implies x^{2b} = (x^{6}) * (x^{d+1})*(x^{b+1})  \implies x^{2b - b - 1} = (x^{6}) * (x^{d+1})\implies x^{b - 1} =  (x^{d+7}) \implies x^{b - 1 - d - 7} =  1  \implies x^{b- d - 8} =  1 \implies b = d+8$$
Therefore $b > d$ option a is correct.

$x= (a^2 + ab)-(ab^2-b)/(2a^2+b^2 -ab)$
Since $b <0$ let $-c = b, c \in \mathbb{R^+}$
$$x= (a^2 - ac)-{(ac^2+c)\over(2a^2+c^2 +ac)}$$
$$x= {(a^2 - ac)(2a^2+c^2 +ac)-(ac^2+c)\over(2a^2+c^2 +ac)} = {2a^4 - a^3c-c^3a-ac^2-c\over(2a^2+c^2 +ac)}$$
By AM-GM,
$$x\ge {2a^4 - 2a^2c^2-ac^2-c\over(2a^2+c^2 +ac)} \ge  {4a^4 - 4a^2c^2-2ac^2-2c\over(5a^2+3c^2)}$$
Let $c = 1$
Therefore,
$$x\ge  {4a^4 - 4a^2-2a-2\over(5a^2+3)}$$
Now let $a = 10$ , then $x \ge 78$
and as you said $x = 0$ for $a = 1, b= -1$.
Therefore option e is correct.
