Question

প্রমাণ কর যে: $\left(\frac{x^{p}}{x^{q}}\right)^{p+q-r}\cdot\left(\frac{x^{q}}{x^{r}}\right)^{q+r-p}$$\cdot\left(\frac{x^{r}}{x^{p}}\right)^{r+p-q}$$=1$

Answer 1

Left Hand Side,

$\left(\frac{x^{p}}{x^{q}}\right)^{p+q-r}\cdot\left(\frac{x^{q}}{x^{r}}\right)^{q+r-p}$$\cdot\left(\frac{x^{r}}{x^{p}}\right)^{r+p-q}$

$=\left(x^{p-q}\right)^{p+q-r}\cdot\left(x^{q-r}\right)^{q+r-p}$$\cdot\left(x^{r-p}\right)^{r+p-q}$

$=x^{\left(p-q\right)\left(p+q-r\right)}\cdot x^{\left(q-r\right)\left(q+r-p\right)}$$\cdot x^{\left(r-p\right)\left(r+p-q\right)}$

$=x^{\left(p-q\right)\left(p+q\right)-r\left(p-q\right)}\cdot x^{\left(q-r\right)\left(q+r\right)-p\left(q-r\right)}$$\cdot x^{\left(r-p\right)\left(r+p\right)-q\left(r-p\right)}$

$=x^{p^2-q^2-pr+qr}\cdot x^{q^2-r^2-pq+pr}\cdot x^{r^2-p^2-qr+pr}$

$=x^{(p^2-q^2-pr+qr)+(q^2-r^2-pq+pr)+(r^2-p^2-qr+pr)}$

$=x^{p^2-q^2-pr+qr+q^2-r^2-pq+pr+r^2-p^2-qr+pr}$

$=x^{0}$

$=1$

$=$ Right Hand Side [Proved]

Comments

Rules Applied:

• $\frac{a^{m}}{a^{n}}=a^{m-n}$
• $\left(a^{m}\right)^{n}=a^{mn}$
• $a^{m} \cdot a^{n}=a^{m+n}$
• $a^{0}=1$