How Can You Factor the Trigonometric Expression sin^3(x)-cos^3(x)?

AI Thread Summary
To factor the expression sin^3(x) - cos^3(x), one can use the identity sin^2(x) + cos^2(x) = 1. The expression can be rewritten as (sin(x) - cos(x))(sin^2(x) + sin(x)cos(x) + cos^2(x)). The denominator sin(x) - cos(x) cancels out, leading to the simplified form of sin^2(x) + sin(x)cos(x) + cos^2(x). There is some confusion about whether the result should include a term with sin(x)cos(x). The discussion emphasizes the importance of factoring and using trigonometric identities effectively.
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Homework Statement



sin^3(x)-cos^3(x)
sin(x) - cos(x)

equals

1 + sin(x) + cos(x)

Homework Equations


Not sure :/


The Attempt at a Solution


Not sure where to even start.
 
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Are you sure that it's 1+sin(x)+cos(x) and not 1+sin(x)cos(x)?

Use the identity sin^2(x)+cos^2(x)=1.
 
Oh yeah, that was it.
 
sin^3(x)-cos^3(x)
-----------------
sin(x) - cos(x)

Try factoring the numerator, it may help you.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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