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Answer 1 · 2017-12-22T12:40:06

#sin^3x+cos^3x#

#=(sinx+cosx)^3-3sinxcosx(sinx+cosx)#

#=(1/3)^3-3*sinxcosx*1/3#

#=1/27-1/2*2sinxcosx#

#=1/27-1/2*[(sinx+cosx)^2-1]#

#=1/27-1/2*[(1/3)^2-1]#

#=1/27-1/2*[1/9-1]#

#=1/27-1/2*(-8/9)#

#=1/27+4/9#

#=1/27+12/27=13/27#

Answer 2 · 2017-12-22T12:40:58

#13/27#

Using the fact

#(a^3+b^3)/(a+b)=a^2-a b+b^2# we have

#sin^3x+cos^3x = 1/3(1-sinx cosx)#

but

#(sinx+cosx)^2= 1+2sinxcosx=1/9# or

#sinx cosx = -4/9# and finally

#sin^3x+cos^3x = 1/3(1+4/9) = 13/27#