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Surds and Indices Practice Test 2

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Question : 9 of 17

Marks: +1, -0

\left(\frac{x^b}{x^c}\right)^{b+c-a}\cdot\left(\frac{x^c}{x^a}\right)^{c+a-b}\cdot\left(\frac{x^a}{x^b}\right)^{a+b-c}=?

$x^{a+b+c}$
$x^{ab+bc+ca}$
1
$x^{abc}$

Solution:

Concept:

Use laws of indices:

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

and

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

.

Explanation:

Simplify each bracket first:

\left(\frac{x^b}{x^c}\right)^{b+c-a}=(x^{b-c})^{b+c-a}

.

\left(\frac{x^c}{x^a}\right)^{c+a-b}=(x^{c-a})^{c+a-b}

.

\left(\frac{x^a}{x^b}\right)^{a+b-c}=(x^{a-b})^{a+b-c}

.

Now multiply powers by adding exponents:

=x^{(b-c)(b+c-a)+(c-a)(c+a-b)+(a-b)(a+b-c)}

.

Expand each product:

(b-c)(b+c-a)=b^2-ab-c^2+ac

.

(c-a)(c+a-b)=c^2-bc-a^2+ab

.

(a-b)(a+b-c)=a^2-ac-b^2+bc

.

Add them:

b^2-ab-c^2+ac+c^2-bc-a^2+ab+a^2-ac-b^2+bc=0

.

So the expression becomes

x^0=1

.

Answer:

1

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