Question

The simplified value of $\frac{x^{a+b}.x^{b+c}.x^{c+a}}{(x^a.x^b.x^c{)}^2}$ is:

• A. $x^2$
• B. $x^{a+b+c}$
• C. $x^{abc}$
• D. $x^0$

Answer 1

The correct option is D $x^0$

The given expression is:

$\frac{x^{a+b}.x^{b+c}.x^{c+a}}{{\left(x^a.x^b.x^c\right)}^2}$

We have, $x^m\times x^n=x^{m+n}$ and similarly,
$x^m\times x^n\times x^p=x^{m+n+p}$

In the given question,
$m=a+b$, $n=b+c$ and $p=c+a$.

Using the property we get,
$\frac{x^{a+b}.x^{b+c}.x^{c+a}}{{\left(x^a.x^b.x^c\right)}^2}$

Also, $(x^m{)}^n=x^{m\times n}$.

Applying this property we get, given expression is,
$=\frac{x^{a+b+b+c+c+a}}{x^{2a+2b+2c}}$
$=\frac{x^{2(a+b+c)}}{x^{2(a+b+c)}}$

Now,
$\frac{x^m}{x^n}=x^{m-n}$

Using this property we get the expression as
$=x^{2a+2b+2c-2a-2b-2c}$
$=x^0$

Hence, option D is correct