Rewriting Brackets: How to Factor Out the s & Distribute Power of 6

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SUMMARY

The discussion focuses on the mathematical process of factoring out 's' and distributing the power of 6 in the expression \(\frac{1}{2}s + \frac{1}{2}s^2\). Participants clarify that the expression can be rewritten as \(\frac{1}{2}s(1+s)\), leading to \(\left(\frac{1}{2}s + \frac{1}{2}s^2\right)^6 = \frac{1}{64}s^6(1+s)^6\). This transformation utilizes the property \((ab)^n = a^n b^n\) to simplify the expression without direct multiplication.

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umzung
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Homework Statement
How does one bracket become the other?
Relevant Equations
See the attempt at a solution.
How does
Screenshot 2019-11-26 22.17.43.png
become
Screenshot 2019-11-26 22.17.54.png
?

I can see the s has been factored out and the power of 6 distributed, but how do we know this happens, short of multiplying out the brackets?
 
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First factor the s in the top expression. Then what happens with the power 6? I think it will be clearer then.
 
umzung said:
Homework Statement: How does one bracket become the other?
Homework Equations: See the attempt at a solution.

How does View attachment 253352become View attachment 253353?

I can see the s has been factored out and the power of 6 distributed, but how do we know this happens, short of multiplying out the brackets?
Just factor the expression inside the parentheses.
##\frac 1 2 s + \frac 1 2 s^2 = s( \frac 1 2 + \frac 1 2 s)##.

Then use the property that ##(ab)^n = a^nb^n##
 
Mark44 said:
Just factor the expression inside the parentheses.
##\frac 1 2 s + \frac 1 2 s^2 = s( \frac 1 2 + \frac 1 2 s)##.

Then use the property that ##(ab)^n = a^nb^n##
That's clearer, thanks.
 
I would actually go a little further than that. \frac{1}{2}s+ \frac{1}{2}s^2= \frac{1}{2}s(1+ s) so that \left(\frac{1}{2}s+ \frac{1}{2}s^2\right)^6= \frac{1}{64}s^6(1+ s)^6.
 

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