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Surds and Indices Practice Test 2

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Question : 3 of 17

Marks: +1, -0

\left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}} \times \left(\frac{x^c}{x^a}\right)^{\frac{1}{ac}} \times \left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}}

is equivalent to:

$x^{b^2 - ac}$
$\frac{1}{x^{ab + b^2 - c^2}}$
$x^{a^3 + b^3 + c^3 - abc}$
1

Solution:

Concept:

Use laws of indices:

\frac{x^m}{x^n}=x^{m-n}

and

\left(x^m\right)^n=x^{mn}

.

Explanation:

First factor:

\left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}}=\left(x^{b-c}\right)^{\frac{1}{bc}}=x^{\frac{1}{c}-\frac{1}{b}}

.

Second factor:

\left(\frac{x^c}{x^a}\right)^{\frac{1}{ac}}=x^{\frac{1}{a}-\frac{1}{c}}

.

Third factor:

\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}}=x^{\frac{1}{b}-\frac{1}{a}}

.

Multiply them: exponents add.

x^{\frac{1}{c}-\frac{1}{b}+\frac{1}{a}-\frac{1}{c}+\frac{1}{b}-\frac{1}{a}}

All terms cancel, so exponent is

0

.

Thus, the expression becomes

x^0=1

.

Answer:

1

, which is option D.

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