Question 993815
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Use this equality:


{{{x^3 + y^3}}} = {{{(x+y)}}}.{{{(x^2 - xy + y^2)}}}.


It is the standard factorization and it is described in any systematic Algebra textbook. &nbsp;See, &nbsp;for example, &nbsp;the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/Polynomials-and-rational-expressions/The-cube-of-a-sum-formula.lesson>The cube of the sum formula</A>&nbsp; in this site. 


Therefore, &nbsp;{{{(x^3 + y^3)/(x+y)}}} = {{{x^2 - xy + y^2}}}.


Now substitute &nbsp;x = sin(A), &nbsp;y = cos(A) &nbsp;into this formula. &nbsp;You will have 


{{{(sin^3(A) + cos^3(A))/(sin(A)+cos(A))}}} = {{{sin^2(A) - sin(A)*cos(A)  + cos^2(A)}}} = {{{1 - sin(A)*cos(A)}}}.


We took into account that &nbsp;{{{sin^2(A) + cos^2(A)}}} = {{{1}}}.