The problem gives an equation $\frac{2x}{a^3 - 8} - \frac{a}{a^2 + 2a + 4} = \frac{x-1}{a-2}$, where $a$ is a real number. We need to: a) Solve the equation for $x$. b) Find the values of $a$ for which the solution is positive. c) Find the value of $a$ for which the solution is $x = 0$.

1. Problem Description

The problem gives an equation $\frac{2x}{a^3 - 8} - \frac{a}{a^2 + 2a + 4} = \frac{x-1}{a-2}$, where $a$ is a real number.

We need to:

a) Solve the equation for $x$.

b) Find the values of $a$ for which the solution is positive.

c) Find the value of $a$ for which the solution is $x = 0$.

2. Solution Steps

a) Solve for $x$:

First, we factor $a^3 - 8$ as a difference of cubes:

$a^3 - 8 = a^3 - 2^3 = (a-2)(a^2 + 2a + 4)$

Now the equation becomes:

$\frac{2x}{(a-2)(a^2 + 2a + 4)} - \frac{a}{a^2 + 2a + 4} = \frac{x-1}{a-2}$

Multiply both sides by $(a-2)(a^2 + 2a + 4)$:

$2x - a(a-2) = (x-1)(a^2 + 2a + 4)$

$2x - a^2 + 2a = x(a^2 + 2a + 4) - (a^2 + 2a + 4)$

$2x - a^2 + 2a = a^2x + 2ax + 4x - a^2 - 2a - 4$

$2x - a^2x - 2ax - 4x = - a^2 - 2a - 4 + a^2 - 2a$

$-a^2x - 2ax - 2x = -4a - 4$

$x(-a^2 - 2a - 2) = -4(a+1)$

$x(a^2 + 2a + 2) = 4(a+1)$

$x = \frac{4(a+1)}{a^2 + 2a + 2}$

b) Find $a$ such that $x > 0$:

We need to find the values of $a$ for which $\frac{4(a+1)}{a^2 + 2a + 2} > 0$.

Since $a^2 + 2a + 2 = (a+1)^2 + 1$, $a^2 + 2a + 2 > 0$ for all real $a$.

Thus, we only need to consider the numerator $4(a+1) > 0$, which implies $a+1 > 0$, so $a > -1$.

c) Find $a$ such that $x = 0$:

We need to find the values of $a$ for which $x = \frac{4(a+1)}{a^2 + 2a + 2} = 0$.

Since $a^2 + 2a + 2 \ne 0$, we must have $4(a+1) = 0$, which means $a+1 = 0$, so $a = -1$.

3. Final Answer

a) $x = \frac{4(a+1)}{a^2 + 2a + 2}$

b) $a > -1$

c) $a = -1$