Solving Exponent Laws: (p^2q+ pq^3)^3 / p^3q^4

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Homework Help Overview

The problem involves simplifying the expression (p^2q + pq^3)^3 / p^3q^4, which relates to exponent laws and algebraic manipulation.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses confusion about how to simplify the expression and seeks guidance. One participant suggests factoring out terms in the numerator and applying exponent rules, while another participant questions the approach of expanding the expression.

Discussion Status

Some participants have provided steps that lead toward a simplified form of the expression, indicating a productive direction in the discussion. However, the original poster's understanding of the process remains a focus, with no explicit consensus reached on the best approach.

Contextual Notes

The original poster mentions being told a specific answer, which may influence their understanding and approach to the problem. There is also an indication of differing methods being considered, such as factoring versus expanding.

majinknight
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Hi, i am having a little trouble with a couple of math questions, i thought i was doing them right but cannot get the answer. Ok the questions are:

(p^2q+ pq^3)^3 / p^3q^4

Now I do not know what to do, the answer i was told is suppose to be:

(p+q^2)^3 / q

I just don't know how to get that, if someone could walk me through this one question it would be appreciated.
 
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you start with:

[tex]\frac{(p^2q+pq^3)^3}{p^3q^4}[/tex]

Factor out pq in the numerator:

[tex]\frac{(pq(p+q^2))^3}{p^3q^4}[/tex]

same as:

[tex]\frac{(pq)^3(p+q^2)^3}{p^3q^4}[/tex]

[tex]\frac{p^3q^3(p+q^2)^3}{p^3q^4}[/tex]

apply division rule for exponents:

[tex]\frac{(p+q^2)^3}{q}[/tex]
 
Oh my god, thank you, i completely understand know. I had sort of done what you had but instead did not factor and so it just kept getting more complicated and confusing me. Thank you so much.
 
What did you do? expand it?, if you had you will get the same just with the numerador expanded, but it will be more work.
 

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