Question

prove that ; (x power b upon x power c) whole power a ( x power c upon x power a ) whole power b ( x power a upon x power b ) = 1

Answer 1

${\huge{\underline{\underline{\sf {\mathfrak{Answer}}}}}}$

${\underline{\underline{\rm {To \ prove:}}}}$

• $\left( \dfrac{x^b}{x^c} \right)^a \ \left( \dfrac{x^c}{x^a}\right)^b \ \left( \dfrac{x^a}{x^b}\right)^c = 1$

${\underline{\underline{\rm {Solving \ LHS:}}}}$

• $\left( \dfrac{x^b}{x^c} \right)^a \ \left( \dfrac{x^c}{x^a}\right)^b \ \left( \dfrac{x^a}{x^b}\right)^c$

• $( x^{b - c})^a (x^{c - a})^b (x^{a-b})^c$

• $x^{ab - ac} \ . \ x^{bc - ab} \ . \ x^{ac - bc}$

• $x^{ab - ac + bc - ab + ac - bc}$

• $x^0$

$\boxed{\implies{\boxed{1 = RHS}}}$

Answer 2

${\bold{\huge{\orange{Hèy Fríéñd}}}}$

Here is the Solution..

${\boxed{Step\:By\:Step\: Solution}}$

We Have :-

$(\frac{x^b}{x^c})^a * (\frac{x^c}{x^a})^b * (\frac{x^a}{x^b})^c = 1$

LHS :-

$(\frac{x^{ab}}{x^{ac}}) * (\frac{x^{bc}}{x^{ab}}) * (\frac{x^{ac}}{x^{bc}}) \\ \\ x^{ab-ac} * x^{bc-ab} * x^{ac-bc} \\ \\ x^{ab-ac+bc-ab+ac-bc} \\ \\ x^{ab-ab+bc-bc+ac-ac} \\ \\ x^0 = 1$

_______________

Therefore, LHS = RHS

${\boxed{\bold{\huge{\red{Hence\:Proved}}}}}$

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