Please explain the full answer.....................

1.	Please explain the full answer.....................
Answer» ANSWER: The value of the expression is 0. GIVEN: $( \dfrac{x {}^{a} }{x {}^{b} } ) {}^{a + b} \times ( \dfrac{x {}^{b} }{x {}^{c} } ) {}^{b + c} \times ( \dfrac{x {}^{c} }{x {}^{a} } ) {}^{c + a}$ TO FIND: The value of above EXPRESSION. SOLUTION: $= ( \dfrac{x {}^{a} }{x {}^{b} } ) {}^{a + b} \times ( \dfrac{x {}^{b} }{x {}^{c} } ) {}^{b + c} \times ( \dfrac{x {}^{c} }{x {}^{a} } ) {}^{c + a} \\ = x{}^{(a - b)(a + b)} \times x{}^{(b - c)(b + c)} \times x{}^{(c - a)(c + a)} \\ = x {}^{(a {}^{2} - b {}^{2}) } \times x {}^{(b {}^{2} - c {}^{2} ) } \times x {}^{(c {}^{2} - a {}^{2}) } \\ = x {}^{a {}^{2} - b {}^{2} + b {}^{2} - c {}^{2} + c {}^{2} - a {}^{2} } \\ \\ all \: power \: will \: cancel \: out. \: we \: get \\ = x {}^{0} \\ = 1$ The value of the expression is 0. NOTE: *some IMPORTANT* *FORMULAS*: $\implies \: a {}^{x} \times a {}^{y} = a {}^{ x + y} \\ \implies \: a {}^{x} \div a {}^{y} = a {}^{x - y} \\ \implies \: a {}^{(m) {}^{n} } = a {}^{mn}$

Answer»

ANSWER:

The value of the expression is 0.

GIVEN:

$( \dfrac{x {}^{a} }{x {}^{b} } ) {}^{a + b} \times ( \dfrac{x {}^{b} }{x {}^{c} } ) {}^{b + c} \times ( \dfrac{x {}^{c} }{x {}^{a} } ) {}^{c + a}$

TO FIND:

The value of above EXPRESSION.

SOLUTION:

$= ( \dfrac{x {}^{a} }{x {}^{b} } ) {}^{a + b} \times ( \dfrac{x {}^{b} }{x {}^{c} } ) {}^{b + c} \times ( \dfrac{x {}^{c} }{x {}^{a} } ) {}^{c + a} \\ = x{}^{(a - b)(a + b)} \times x{}^{(b - c)(b + c)} \times x{}^{(c - a)(c + a)} \\ = x {}^{(a {}^{2} - b {}^{2}) } \times x {}^{(b {}^{2} - c {}^{2} ) } \times x {}^{(c {}^{2} - a {}^{2}) } \\ = x {}^{a {}^{2} - b {}^{2} + b {}^{2} - c {}^{2} + c {}^{2} - a {}^{2} } \\ \\ all \: power \: will \: cancel \: out. \: we \: get \\ = x {}^{0} \\ = 1$

The value of the expression is 0.

NOTE:

some IMPORTANT FORMULAS:

$\implies \: a {}^{x} \times a {}^{y} = a {}^{ x + y} \\ \implies \: a {}^{x} \div a {}^{y} = a {}^{x - y} \\ \implies \: a {}^{(m) {}^{n} } = a {}^{mn}$