SOLUTION: Find all real numbers $a$ that satisfy \frac{1}{a^3 + 7} -7 = -\frac{a^2}{a^3 + 7}.

Question 1209051: Find all real numbers $a$ that satisfy
\frac{1}{a^3 + 7} -7 = -\frac{a^2}{a^3 + 7}.
Answer by yurtman(42) (Show Source): You can put this solution on YOUR website!
To solve the equation, we can first multiply both sides by $a^3 + 7$ to eliminate the denominators:
$$1 - 7(a^3 + 7) = -a^2$$
Expanding the left side, we get:
$$1 - 7a^3 - 49 = -a^2$$
Rearranging the terms, we obtain a cubic equation:
$$7a^3 - a^2 - 48 = 0$$
Unfortunately, there's no simple algebraic method to solve cubic equations in general. We can use numerical methods or software tools to find approximate solutions. However, we can try to factor the equation to see if we can find any rational solutions.
By trial and error, we can find that $a = 2$ is a solution. This means that $(a-2)$ is a factor of the cubic polynomial. We can perform polynomial long division to find the other factor:
$$7a^3 - a^2 - 48 = (a-2)(7a^2 + 13a + 24)$$
The quadratic factor $7a^2 + 13a + 24$ has no real roots, as its discriminant is negative.
Therefore, the only real solution to the original equation is:
$$a = 2$$

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