What is the simplified value of the expression \left(\frac{64}{125}\right)^{-\frac{2}{3}}? — State PSC Exam

What is the simplified value of the expression $\left(\frac{64}{125}\right)^{-\frac{2}{3}}$ ?

\frac{25}{16}

The expression

\left(\frac{64}{125}\right)^{-\frac{2}{3}}

is simplified by first converting the negative power to positive by inverting the fraction to

\left(\frac{125}{64}\right)^{\frac{2}{3}}

. Taking the cube root yields

\frac{5}{4}

, and squaring that result gives

\frac{25}{16}

Apply the negative exponent rule

a^{-n} = \frac{1}{a^n}

to invert the fraction

\left(\frac{64}{125}\right)^{-\frac{2}{3}} = \left(\frac{125}{64}\right)^{\frac{2}{3}}

A negative exponent indicates taking the reciprocal of the base.

Express the base numbers as perfect cubes

125 = 5^3

and

64 = 4^3

, so

\frac{125}{64} = \left(\frac{5}{4}\right)^3

Rewriting bases into prime factors with power multiples simplifies fractional exponents.

Apply the power of a power rule

(a^m)^n = a^{m \times n}

\left(\left(\frac{5}{4}\right)^3\right)^{\frac{2}{3}} = \left(\frac{5}{4}\right)^{3 \times \frac{2}{3}} = \left(\frac{5}{4}\right)^2

Multiplying the inner exponent

3

by the outer exponent

\frac{2}{3}

yields

2

Square the fraction

\left(\frac{5}{4}\right)^2 = \frac{25}{16}

Square both the numerator and the denominator to get the final numerical value.

Negative and Fractional Indices Rules

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