Solve: Finding the Numerator in (?)/(24x³-8x²) = 1/(8x²)

Let's examine the problem:

$\frac{?}{24x^3-8x^2}=\frac{1}{8x^2}$

First we'll check that in the fraction's numerator which is on the left side there is an expression that can be factored using factoring out a common factor. We will therefore factor out the largest common factor possible (meaning that the expression left in parentheses cannot be further factored by taking out a common factor):

$\frac{?}{24x^3-8x^2}=\frac{1}{8x^2} \\ \downarrow\\ \frac{?}{8x^2(3x-1)}=\frac{1}{8x^2} \\$In factoring, we used of course the law of exponents:

$\bm{a^{m+n}=a^m\cdot a^n}$

Let's continue solving the problem. Remember the reduction operation of a fraction. Note that in the fraction's numerator both on the right side and on the left side there is the expression:$8x^2$, therefore we don't want to reduce from the fraction's numerator which is on the left side. However, the expression:

$3x-1$,

is not found in the fraction's numerator which is on the right side (which is the fraction after reduction) Therefore we can conclude that this expression needs to be reduced from the fraction's numerator which is on the left side. Hence the missing expression must be none other than:

$3x-1$

Let's verify that this choice gives us the expression which is in the right side: (reduction sign)

$\frac{?}{8x^2(3x-1)}=\frac{1}{8x^2} \\ \downarrow\\ \frac{\textcolor{red}{3x-1}}{8x^2(3x-1)}\stackrel{?}{= }\frac{1}{8x^2} \\ \downarrow\\ \boxed{\frac{1}{8x^2} \stackrel{!}{= }\frac{1}{8x^2} }$

Therefore the following expression:

$3x-1$

is indeed correct.

Which means that the correct answer is answer A.