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Question : 36 of 86

Marks: +1, -0

Assuming that

x, y, z

are positive real numbers and the exponents are all rational numbers, then

\left(\; \frac{x^{a}}{x^{b}} \right)^{a^{2}+a b+b^{2}} \left(\; \frac{x^{b}}{x^{c}} \right)^{b^{2}+b c+c^{2}} \left(\; \frac{x^{c}}{x^{a}} \right)^{c^{2}+a c+a^{2}}

$-1$
1
0
2

Solution:

Concept:

Simplify the expression using exponent rules:

\frac{x^a}{x^b} = x^{a-b}

and

(x^m)^n = x^{mn}

.

Explanation:

First factor:

\left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2} = \left(x^{a-b}\right)^{a^2+ab+b^2} = x^{(a-b)(a^2+ab+b^2)} = x^{a^3 - b^3}

.

Second factor:

\left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2} = x^{(b-c)(b^2+bc+c^2)} = x^{b^3 - c^3}

.

Third factor:

\left(\frac{x^c}{x^a}\right)^{c^2+ac+a^2} = x^{(c-a)(c^2+ac+a^2)} = x^{c^3 - a^3}

.

Multiply the exponents:

(a^3 - b^3) + (b^3 - c^3) + (c^3 - a^3) = 0

.

Thus, the product is

x^0 = 1

(since

x>0

).

Answer:

1

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