Question

solve the attachment given:☝️☝️☝️☝️☝️​

Answer 1

$\huge\bold\blue{Question}$

$\bold{(x^{a-b})^{a+b}×(x^{b-c})^{b+c}×(x^{c-a})^{c+a}=1}$

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Simplified \ form \ is}$

$\bold{x^{-a^{2}c^{2}(c^{2}-a^{2})-a^{2}b^{2}(a^{2}-b^{2})-b^{2}c^{2}(b^{2}-c^{2})}=1}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{(x^{a-b})^{a+b}×(x^{b-c})^{b+c}×(x^{c-a})^{c+a}=1}$

$\bold{x^{a^{2}-b^{2}}×x^{b^{2}-c^{2}}×x^{c^{2}-a^{2}}=1}$

$\bold{x^{(a^{2}-b^{2})×(b^{2}-c^{2})×(c^{2}-a^{2})}=1}$

$\bold{x^{(a^{2}b^{2}-a^{2}c^{2}-b^{4}+b^{2}c^{2})(c^{2}-a^{2})}=1}$

$\bold{x^{c^{2}a^{2}b^{2}-a^{2}c^{4}-b^{4}c^{2}+b^{2}c^{4}-a^{4}b^{2}+a^{4}c^{2}+a^{2}b^{4}-a^{2}b^{2}c^{2}}=1}$

$\bold{x^{-a^{2}c^{4}-b^{4}c^{2}+b^{2}c^{4}-a^{4}b^{2}+a^{4}c^{2}+a^{2}b^{4}}=1}$

$\bold{x^{-a^{2}c^{4}+a^{4}c^{2}-a^{4}b^{2}+a^{2}b^{4}-b^{4}c^{2}+b^{2}c^{4}}=1}$

$\bold{x^{-a^{2}c^{2}(c^{2}-a^{2})-a^{2}b^{2}(a^{2}-b^{2})-b^{2}c^{2}(b^{2}-c^{2})}=1}$