Question

plz solve this question ​

Answer 1

Formula Applied on it :

$\underbrace{\sf Laws\ of\ Exponents}:\\ \\ \sf 1.\ a^x\times a^y=a^{x+y}\\\\\sf 2.\ \dfrac{a^x}{a^y}=a^{x-y}\\ \\ \sf 3.\ (a^x)^y=a^{xy}\\\\\sf 4.\ \sqrt[\sf{n}]{\sf{a}}=a^{\frac{1}{n}}$

Solution :-

$\sf \sqrt[\sf x+y]{\dfrac{\sf a^{x^2}}{\sf a^{y^2}}} \times \sqrt[\sf y+z]{\dfrac{\sf a^{y^2}}{\sf a^{z^2}}}\times \sqrt[\sf z+x]{\dfrac{\sf a^{z^2}}{\sf a^{x^2}}}$

By applying 4th law of exponent ,

$\mapsto\sf ( \dfrac{a^{x^2}}{a^{y^2}})^{\frac{1}{x+y}}\times (\dfrac{a^{y^2}}{a^{z^2}})^{\frac{1}{y+z}}\times (\dfrac{a^{z^2}}{a^{x^2}})^{\frac{1}{z+x}}$

By applying 2nd law of exponent ,

$\mapsto \sf (a^{x^2-y^2})^{\frac{1}{x+y}}\times (a^{y^2-z^2})^{\frac{1}{y+z}}\times (a^{z^2-x^2})^{\frac{1}{z+x}}$

By applying 3rd law of exponent ,

$\mapsto \sf a^{\frac{x^2-y^2}{x+y}}\times a^{\frac{y^2-z^2}{y+z}}\times a^{\frac{z^2-x^2}{z+x}}$

By applying a² - b² = ( a+b )( a-b ) ,

$\mapsto \sf{ a^{\frac{\cancel{(x+y)}(x-y)}{\cancel{x+y}}}\times a^{\frac{\cancel{(y+z)}(y-z)}{\cancel{y+z}}}\times a^{\frac{\cancel{(z+x)}(z-x)}{\cancel{z+x}}}}$

$\mapsto \sf a^{x-y}\times a^{y-z}\times a^{z-x}$

By applying 1st law of exponent ,

$\mapsto \sf a^{[(x-y)+(y-z)+(z-x)]}$

$\mapsto \sf a^{\cancel{x}\cancel{-y}\cancel{+y}\cancel{-z}\cancel{+z}\cancel{-x}}$

$\mapsto \sf a^{\circ}$

$\mapsto \huge{\boxed {\tt {1}}}$

$\therefore{\underline{\pink{\sf{The\:answer\:is\:1.}}}}$