Q. Fully simplify: \( (-2 x^{3} y^{3})^{5} = -32 x^{15} y^{15} \).

Q: How do I fully simplify (-2x^3y^3)^5 ?

Use power-of-a-product and power-of-a-power: (-2)^5 (x^3)^5 (y^3)^5 = -32x^{15}y^{15} .

Q: Why can I distribute the exponent across multiplication?

Because (ab)^n = a^n b^n for integer n; the exponent applies to each factor inside the parentheses.

Q: How do I handle the negative sign in (-2)^5 ?.

(-2)^5 = (-1)^5\cdot 2^5 = -32. An odd exponent keeps the negative sign; an even exponent would make it positive.

Q: How does (x^3)^5 become x^{15} ?.

By the power-of-a-power rule: (x^a)^b = x^{ab}. Here 3\cdot5=15.

Q: Could I expand (-2x^3y^3)^5 by multiplying everything out?

Yes, but expansion is lengthy; the compact simplified form -32x^{15}y^{15} is preferred.

Q: What if the exponent were 0 or even, e.g. (-2x^3y^3)^0 or ^4?.

What if the exponent were 0 or even, e.g. (-2x^3y^3)^0 or ^4?.

Answer

Use the power rules \( (ab)^n = a^n b^n \) and \( (x^m)^n = x^{mn} \):

\[ (-2x^{3}y^{3})^{5} = (-2)^{5}x^{15}y^{15} = -32x^{15}y^{15} \]

Detailed Explanation

Write the original expression: \( (-2x^{3}y^{3})^{5} \).
Apply the power-of-a-product rule: for any factors \(a,b,c\) and integer \(n\), \((abc)^{n}=a^{n}b^{n}c^{n}\). Thus
\[
(-2x^{3}y^{3})^{5}=(-2)^{5}\,(x^{3})^{5}\,(y^{3})^{5}.
\]
Use the power-of-a-power rule: \((a^{m})^{n}=a^{mn}\). So
\[
(x^{3})^{5}=x^{3\cdot 5}=x^{15},\qquad (y^{3})^{5}=y^{3\cdot 5}=y^{15}.
\]
Evaluate the numerical factor \( (-2)^{5} \). Compute successive powers:
\[
(-2)^{2}=4,\quad (-2)^{3}=-8,\quad (-2)^{4}=16,\quad (-2)^{5}=-32.
\]
Therefore \( (-2)^{5}=-32\).
Combine the results:
\[
(-2x^{3}y^{3})^{5}=(-32)\,x^{15}\,y^{15}=-32x^{15}y^{15}.
\]

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Algebra FAQs

How do I fully simplify \( (-2x^3y^3)^5 \)?

Use power-of-a-product and power-of-a-power: \( (-2)^5 (x^3)^5 (y^3)^5 = -32x^{15}y^{15} \).

Why can I distribute the exponent across multiplication?

Because \( (ab)^n = a^n b^n\) for integer \(n\); the exponent applies to each factor inside the parentheses.

How do I handle the negative sign in \( (-2)^5 \)?.

\( (-2)^5 = (-1)^5\cdot 2^5 = -32\). An odd exponent keeps the negative sign; an even exponent would make it positive.

How does \( (x^3)^5 \) become \( x^{15} \)?.

By the power-of-a-power rule: \( (x^a)^b = x^{ab}\). Here \(3\cdot5=15\).

Could I expand \( (-2x^3y^3)^5 \) by multiplying everything out?

Yes, but expansion is lengthy; the compact simplified form \( -32x^{15}y^{15} \) is preferred.

What if the exponent were \(0\) or even, e.g. \( (-2x^3y^3)^0\) or \( ^4\)?.

Do variable values (negative or zero) change the simplification rules?

The algebraic rules stay the same. If a variable is negative, its sign after exponent depends on parity; if any variable equals \(0\) and \( \)exponent \( \gt 0\), the whole product is \(0\).

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