Question

solve this problem with step quickly

Answer 1

mark as brilliant answer

Answer 2

$\mathsf{Given :\;\bigg(\dfrac{x^a}{x^b}\bigg)^{a + b} \times \bigg(\dfrac{x^b}{x^c}\bigg)^{b + c} \times \bigg(\dfrac{x^c}{x^a}\bigg)^{c + a}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\dfrac{a^m}{a^n} = a^{m - n}}}}$

$\mathsf{\implies \dfrac{x^a}{x^b} = x^{a - b}}$

$\mathsf{\implies \dfrac{x^b}{x^c} = x^{b - c}}$

$\mathsf{\implies \dfrac{x^c}{x^a} = x^{c - a}}$

$\mathsf{\implies \big[x^{a - b}\big]^{a + b} \times \big[x^{b - c}\big]^{b + c} \times \big[x^{c - a}\big]^{c + a}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{\big[a^m\big]^n = a^{mn}}}}$

$\mathsf{\implies \big[x\big]^{(a + b)(a - b)} \times \big[x\big]^{(b + c)(b - c)} \times \big[x\big]^{(c + a)(c - a)}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{(P + Q)(P - Q) = P^2 - Q^2}}}$

$\mathsf{\implies \big[x\big]^{a^2 - b^2} \times \big[x\big]^{b^2 - c^2} \times \big[x\big]^{c^2 - a^2}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^m \times a^n = a^{m + n}}}}$

$\mathsf{\implies \big[x\big]^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}}$

$\mathsf{\implies \big[x\big]^{0}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathsf{a^0 = 1}}}$

$\mathsf{\implies \big[x\big]^{0} = 1}$