1Cademy - Simplifying $$-25^{\frac{3}{2}}$$, $$-25^{-\frac{3}{2}}$$, and $$(-25)^{\frac{3}{2}}$$

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Rational Exponent $a^{\frac{m}{n}}$

Example

Simplifying $-25^{\frac{3}{2}}$ , $-25^{-\frac{3}{2}}$ , and $(-25)^{\frac{3}{2}}$

Simplify three expressions that each involve the number $25$ , a rational exponent $\frac{3}{2}$ , and a negative sign — but with the negative sign and exponent sign placed differently in each part. The three variations illustrate how parentheses and the sign of the exponent change both the procedure and the outcome.

ⓐ $-25^{\frac{3}{2}} = -125$ : Without parentheses, only $25$ is the base and the leading negative sign is applied last. Convert the rational exponent to radical form using $a^{\frac{m}{n}} = (\sqrt[n]{a})^m$ : $-25^{\frac{3}{2}} = -(\sqrt{25})^3$ . Evaluate the square root: $\sqrt{25} = 5$ . Raise to the third power: $5^3 = 125$ . Apply the negation: $-125$ .

ⓑ $-25^{-\frac{3}{2}} = -\frac{1}{125}$ : Again, only $25$ is the base — the leading negative sign is separate. The exponent $-\frac{3}{2}$ is negative, so first apply the negative exponent rule $b^{-p} = \frac{1}{b^p}$ : $-25^{-\frac{3}{2}} = -\left(\frac{1}{25^{\frac{3}{2}}}\right)$ . Now convert $25^{\frac{3}{2}}$ to radical form: $\frac{1}{(\sqrt{25})^3}$ . Since $\sqrt{25} = 5$ , evaluate: $\frac{1}{5^3} = \frac{1}{125}$ . Apply the negation: $-\frac{1}{125}$ .

ⓒ $(-25)^{\frac{3}{2}}$ is not a real number: The parentheses make $-25$ the entire base. Convert to radical form: $(-25)^{\frac{3}{2}} = (\sqrt{-25})^3$ . However, there is no real number whose square equals $-25$ , so $\sqrt{-25}$ does not exist in the real numbers. Therefore the entire expression is not a real number.

Parts ⓐ and ⓑ both produce real (negative) results because the base being raised to the $\frac{3}{2}$ power is the positive number $25$ , and the negation is applied afterward. Part ⓒ fails because placing a negative number inside the parentheses forces the computation to take the square root of a negative number, which is undefined in the real number system. This contrasts with the cube-root case $(-64)^{\frac{1}{3}} = -4$ , which is real because odd roots of negative numbers exist.

Updated 2026-05-01

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