Question

If $x$ is a real number and $p,q,r$are whole numbers, then prove that ${\left(\frac{x^p}{x^q}\right)}^{p+q}\times {\left(\frac{x^r}{x^p}\right)}^{r+p}\times {\left(\frac{x^q}{x^r}\right)}^{q+r}=1$

Answer 1

Prove that ${\left(\frac{x^p}{x^q}\right)}^{p+q}\times {\left(\frac{x^r}{x^p}\right)}^{r+p}\times {\left(\frac{x^q}{x^r}\right)}^{q+r}=1$

$\begin{array}{l}{\left(\frac{x^p}{x^q}\right)}^{p+q}\times {\left(\frac{x^r}{x^p}\right)}^{r+p}\times {\left(\frac{x^q}{x^r}\right)}^{q+r}=1\\ =L\mathrm{HS}\\ ={\left(\frac{x^p}{x^q}\right)}^{p+q}\times {\left(\frac{x^r}{x^p}\right)}^{r+p}\times {\left(\frac{x^q}{x^r}\right)}^{q+r}\\ ={\left(x^{p-q}\right)}^{p+q}\times {\left(x^{r-p}\right)}^{r+p}\times {\left(x^{q-r}\right)}^{q+r}\left[\because a^{m-n}=\frac{a^m}{a^n}\right]\\ ={\left(x\right)}^{p^2-q^2}\times {\left(x\right)}^{r^2-p^2}\times {\left(x\right)}^{q^2-r^2}\left(\because {\left(a^m\right)}^n=a^{mn}\right)\\ ={\left(x\right)}^{\left(p^2-q^2+r^2-p^2+q^2-r^2\right)}\\ =x^0\left(\because a^0=1\right)\\ =1\end{array}$

Hence ,it is proved that ${\left(\frac{x^p}{x^q}\right)}^{p+q}\times {\left(\frac{x^r}{x^p}\right)}^{r+p}\times {\left(\frac{x^q}{x^r}\right)}^{q+r}=1$