SOLUTION: I have been working on this math problem and I can't figure it out. I was wondering if someone could help me? Please and Thank You!!! I would deeply appreciate it!! Factoring Qu

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Question 166270This question is from textbook Algebra and Trigonometry Structure and Method book 2
: I have been working on this math problem and I can't figure it out. I was wondering if someone could help me? Please and Thank You!!! I would deeply appreciate it!!
Factoring Quadratic Polynomials
Factor completely. If the polynomial is prime, say so.
x^6-64y^6 This question is from textbook Algebra and Trigonometry Structure and Method book 2

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E6-64y%5E6 Start with the given expression.


%28x%5E3%29%5E2-64y%5E6 Rewrite x%5E6 as %28x%5E3%29%5E2.


%28x%5E3%29%5E2-%288y%5E3%29%5E2 Rewrite 64y%5E6 as %288y%5E3%29%5E2.


Notice how we have a difference of squares. So let's use the difference of squares formula A%5E2-B%5E2=%28A%2BB%29%28A-B%29 to factor the expression:


%28x%5E3%2B8y%5E3%29%28x%5E3-8y%5E3%29 Factor the expression using the difference of squares.


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Now let's factor x%5E3%2B8y%5E3


x%5E3%2B8y%5E3 Start with the given expression.


%28x%29%5E3%2B%282y%29%5E3 Rewrite x%5E3 as %28x%29%5E3. Rewrite 8y%5E3 as %282y%29%5E3.


%28x%2B2y%29%28%28x%29%5E2-%28x%29%282y%29%2B%282y%29%5E2%29 Now factor by using the sum of cubes formula. Remember the sum of cubes formula is A%5E3%2BB%5E3=%28A%2BB%29%28A%5E2-AB%2BB%5E2%29


%28x%2B2y%29%28x%5E2-2xy%2B4y%5E2%29 Multiply


So x%5E3%2B8y%5E3 factors to %28x%2B2y%29%28x%5E2-2xy%2B4y%5E2%29.


In other words, x%5E3%2B8y%5E3=%28x%2B2y%29%28x%5E2-2xy%2B4y%5E2%29



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Now let's factor x%5E3-8y%5E3


x%5E3-8y%5E3 Start with the given expression.


%28x%29%5E3-%282y%29%5E3 Rewrite x%5E3 as %28x%29%5E3. Rewrite 8y%5E3 as %282y%29%5E3.


%28x-2y%29%28%28x%29%5E2%2B%28x%29%282y%29%2B%282y%29%5E2%29 Now factor by using the difference of cubes formula. Remember the difference of cubes formula is A%5E3-B%5E3=%28A-B%29%28A%5E2%2BAB%2BB%5E2%29


%28x-2y%29%28x%5E2%2B2xy%2B4y%5E2%29 Multiply


So x%5E3-8y%5E3 factors to %28x-2y%29%28x%5E2%2B2xy%2B4y%5E2%29.


In other words, x%5E3-8y%5E3=%28x-2y%29%28x%5E2%2B2xy%2B4y%5E2%29

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Answer:

So x%5E6-64y%5E6 completely factors to %28x%2B2y%29%28x-2y%29%28x%5E2-2xy%2B4y%5E2%29%28x%5E2%2B2xy%2B4y%5E2%29 (I just rearranged the factors a bit)


Note: the order of the factors does not matter