We can rewrite this problem so that it's a simpler multiplication problem. Take the bottom of the express (\(\displaystyle abc\)) and write its reciprocal, then multiply it by the fraction in the numerator, like so:

\(\displaystyle \frac{\frac{a^2b^3c}{a^5b^7c^4}}{abc} = \frac{a^2b^3c}{a^5b^7c^4} \cdot \frac{1}{abc}\)

Now, we can simplify using cross-division. 

\(\displaystyle \frac{a^2b^3c}{a^5b^7c^4} \cdot \frac{1}{abc} = \frac{ab^2}{a^5b^7c^4}\)

Now, we can simplify this expression. When we divide these terms, we subtract their exponents.

\(\displaystyle \frac{ab^2}{a^5b^7c^4} = \frac{1}{a^4b^5c^4}\)

Separate each group of variables to simplify.

\(\displaystyle \frac{a^{4}}{a^{6}} =\) \(\displaystyle \frac{a\times a\times a\times a}{a \times a \times a \times a \times a \times a}\) \(\displaystyle = \frac{1}{a^{2}}\)

\(\displaystyle \frac{b^{-2}}{b^{3}}=\) \(\displaystyle \frac{1}{b^{2}}\times\frac{1}{b^{3}}=\) \(\displaystyle \frac{1}{b\times b\times b\times b\times b}=\frac{1}{b^{5}}\)

\(\displaystyle \frac{c^{5}}{c^{7}} =\) \(\displaystyle \frac{c\times c\times c\times c\times c}{c\times c\times c\times c\times c\times c\times c}\) \(\displaystyle = \frac{1}{c^{2}}\)

\(\displaystyle \frac{d^{14}}{d^{-7}}=\frac{d^{14}}{1}\times \frac{d^{7}}{1}\)\(\displaystyle = \frac{d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d\times d}{1}\)

\(\displaystyle = \frac{d^{21}}{1}\)

\(\displaystyle \frac{e^{10}}{e^{8}}=\frac{e\times e\times e\times e\times e\times e\times e\times e\times e\times e}{e\times e\times e\times e\times e\times e\times e\times e}\) \(\displaystyle = \frac{e^{2}}{1}\)

\(\displaystyle \frac{1}{a^{2}}\times \frac{1}{b^{5}}\times \frac{1}{c^{2}}\times \frac{d^{21}}{1}\times \frac{e^{2}}{1}= \frac{d^{21}e^{2}}{a^{2}b^{5}c^{2}}\)

Take the reciprocal of both expressions:

\(\displaystyle \frac{1}{u+1} = t\)

\(\displaystyle u+1 = \frac{1}{t}\)

Subtract 5 from both sides:

\(\displaystyle u+1-5 = \frac{1}{t}-5\)

\(\displaystyle u-4 = \frac{1}{t}-5\)

Rewrite the expression at left and simplify it:

\(\displaystyle u-4 = \frac{1}{t}-\frac{5t}{t}\)

\(\displaystyle u-4 = \frac{1-5t}{t}\)

Take the reciprocal of both expressions:

\(\displaystyle \frac{1}{u-4} = \frac{t}{1-5t}\)

Square both expressions

\(\displaystyle (\sqrt{x+ 1} )^{2}= t^{2}\)

\(\displaystyle x+1 = t^{2}\)

Subtract 8 from both sides:

\(\displaystyle x+1-8 = t^{2} -8\)

\(\displaystyle x-7= t^{2} - 8\)

Take the positive square root of both sides:

\(\displaystyle \sqrt{x-7}= \sqrt{t^{2} -8}\)

Square both expressions

\(\displaystyle (\sqrt{x+ 1} )^{2}= t^{2}\)

\(\displaystyle x+1 = t^{2}\)

Add 6 to both sides:

\(\displaystyle x+1+ 6 = t^{2} + 6\)

\(\displaystyle x+7= t^{2} + 6\)

Take the square root of both sides:

\(\displaystyle \sqrt{x+7}= \sqrt{t^{2} + 6}\)

\(\displaystyle (x+1) ^{2} = t\), so, taking the square root of both sides:

\(\displaystyle x+1 = \pm\sqrt{ t}\)

\(\displaystyle x\) is positive, so \(\displaystyle x+1\) is as well; therefore, 

\(\displaystyle x+1 = \sqrt{ t}\)

Add 4 to both sides:

\(\displaystyle x+1 + 4 = \sqrt{ t} + 4\)

\(\displaystyle x+5 =\sqrt{ t} + 4\)

Square both sides, and apply the binomial square pattern to the right expression:

\(\displaystyle (x+5 )^{2}= (\sqrt{ t} + 4)^{2}\)

\(\displaystyle = (\sqrt{ t})^{2}+ 2\cdot4 \cdot (\sqrt{ t}) +4 ^{2}\)

\(\displaystyle =t + 8\sqrt{ t} +16\)

Take the reciprocal of both expressions:

\(\displaystyle \frac{1}{u+1} = t\)

\(\displaystyle u+1 = \frac{1}{t}\)

Add 4 to both sides:

\(\displaystyle u+1+ 4 = \frac{1}{t}+ 4\)

\(\displaystyle u+5 = \frac{1}{t}+ 4\)

Rewrite the expression at left and simplify it:

\(\displaystyle u+5 = \frac{1}{t}+ \frac{4t}{t}\)

\(\displaystyle u+5 = \frac{1+4t}{t}\)

Take the reciprocal of both expressions:

\(\displaystyle \frac{1}{u+5} = \frac{t}{1+4t}\)

To simplify this expression, you must combine like terms. You should first use the distributive property and multiply -4 by x² and -4 by 3.

x³ - 4x² -12 + 15

You can then add -12 and 15, which equals 3.

You now have x³ - 4x² + 3 and are finished. Just a reminder that x³ and 4x² are not like terms as the x’s have different exponents.

A negative exponent in the numerator of a fraction can be rewritten with a positive exponent in the denominator. The same is true for a negative exponent in the denominator. Thus, \(\displaystyle \frac{a^{-2}}{b^{-3}} =\frac{b^{3}}{a^{2}}\).

When \(\displaystyle \frac{a^{2}}{b^{3}}\) is multiplied by \(\displaystyle \frac{b^{3}}{a^{2}}\), the numerators and denominators cancel out, and you are left with 1.

Another way to represent this question is:

\(\displaystyle \begin{matrix} & \ \ AB\\ +& \ \ BA\\ & 1A2 \end{matrix}\)

In the one's column, \(\displaystyle B\) and \(\displaystyle A\) add to produce a number with a two in the one's place. In the ten's column, we can see that a one must carry in order to get a digit in the hundred's place. Together, we can combine these deductions to see that the sum of \(\displaystyle A\) and \(\displaystyle B\) must be twelve (a one in the ten's place and a two in the one's place).

In the one's column: \(\displaystyle B+A=12\)

The one carries to the ten's column.

In the ten's column: \(\displaystyle (1)+A+B=(1)+12=13\)

The three goes into the answer and the one carries to the hundred's place. The final answer is 132. From this, we can see that \(\displaystyle {A=3}\) because \(\displaystyle 1A2=132\).

Using this information, we can solve for \(\displaystyle B\).

\(\displaystyle A+B=12\ \text{and}\ A=3\)

\(\displaystyle 3+B=12\)

\(\displaystyle B=9\)

You can check your answer by returning to the original addition and plugging in the values of \(\displaystyle A\) and \(\displaystyle B\).

\(\displaystyle AB=39\ \text{and}\ BA=93\)

\(\displaystyle AB+BA=39+93=132\)