Question

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

Answer 1

Left Hand Side,

$\left(\frac{x^{a}}{x^{b}}\right)^{\frac{1}{ab}}\cdot\left(\frac{x^{b}}{x^{c}}\right)^{\frac{1}{bc}}\cdot\left(\frac{x^{c}}{x^{a}}\right)^{\frac{1}{ca}}$

$=\left(x^{a-b}\right)^{\frac{1}{ab}}\cdot\left(x^{b-c}\right)^{\frac{1}{bc}}\cdot\left(x^{c-a}\right)^{\frac{1}{ca}}$

$=x^{\frac{a-b}{ab}}\cdot x^{\frac{b-c}{bc}}\cdot x^{\frac{c-a}{ca}}$

$=x^{\frac{a-b}{ab}+\frac{b-c}{bc}+\frac{c-a}{ca}}$

$=x^{\frac{c(a-b)+a(b-c)+b(c-a)}{abc}}$

$=x^{\frac{ac-bc+ab-ac+bc-ab}{abc}}$

$=x^{\frac{0}{abc}}$

$=x^0$

$=1$

$=$ Right Hand Side [Proved]

Comments

Rules Applied:

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