Question

প্রমাণ কর যে: $\left(\frac{x^{a}}{x^{b}}\right)^{a+b}\cdot\left(\frac{x^{b}}{x^{c}}\right)^{b+c}\cdot\left(\frac{x^{c}}{x^{a}}\right)^{c+a}=1$

Answer 1

Left Hand Side
$=\left(\frac{x^{a}}{x^{b}}\right)^{a+b}\cdot\left(\frac{x^{b}}{x^{c}}\right)^{b+c}\cdot\left(\frac{x^{c}}{x^{a}}\right)^{c+a}$

$=\left(x^{a-b}\right)^{a+b}\cdot\left(x^{b-c}\right)^{b+c}\cdot\left(x^{c-a}\right)^{c+a}$

$=x^{\left(a-b\right)\left(a+b\right)}\cdot x^{\left(b-c\right)\left(b+c\right)}\cdot x^{\left(c-a\right)\left(c+a\right)}$

$=x^{\left(a^2-b^2\right)}\cdot x^{\left(b^2-c^2\right)}\cdot x^{\left(c^2-a^2\right)}$

$=x^{\left(a^2-b^2\right)+\left(b^2-c^2\right)+\left(c^2-a^2\right)}$

$=x^{a^2-b^2+b^2-c^2+c^2-a^2}$

$=x^0$

$=1$

$=$ Right Hand Side [Proved]

Comments

Rules Applied:

• $\frac{a^{m}}{a^{n}}=a^{m-n}$
• $\left(a^{m}\right)^{n}=a^{mn}$
• $(a+b)(a-b)=a^2-b^2$
• $a^{0}=1$