Test Index

Indices

Question : 22 of 86

Marks: +1, -0

\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} x \left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}} \times \left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}

find the value of

Solution:

Concept:

Simplify the product of exponential terms using exponent rules and the property that the sum of cyclic fractions vanishes.

Explanation:

The expression is

\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}} \cdot \left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}} \cdot \left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}

(assuming the middle

x

is a multiplication sign).

Multiply exponents:

x^{\frac{1}{(a-b)(a-c)}} \cdot x^{\frac{1}{(b-c)(b-a)}} \cdot x^{\frac{1}{(c-a)(c-b)}}

The total exponent is

S = \frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-b)}

Rewrite denominators using

(b-a)=-(a-b)

(c-a)=-(a-c)

(c-b)=-(b-c)

\frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(- (a-b))} + \frac{1}{(-(a-c))(-(b-c))}

= \frac{1}{(a-b)(a-c)} - \frac{1}{(a-b)(b-c)} + \frac{1}{(a-c)(b-c)}

Combine over common denominator

(a-b)(a-c)(b-c)

\frac{(b-c) - (a-c) + (a-b)}{(a-b)(a-c)(b-c)} = \frac{b-c - a + c + a - b}{(a-b)(a-c)(b-c)} = \frac{0}{(a-b)(a-c)(b-c)} = 0

Thus

S=0

, so the product equals

x^0 = 1

Answer:

1

Go to Question:

Prev Question

Next Question