Question

sin cube theta + cos cube theta upon sin theta + cos theta + sin theta cos theta = 1

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Answer 1

Let theta = @

(sin^3@+cos^3@)/(sin@+cos@+sin@cos@)=1

Transposing

(sin@+cos@)(sin^2@+cos^2@-sin@cos@)=(sin@+cos@+sin@cos@)

=> (sin@+cos@)(1-sin@cos@)=(sin@+cos@+sin@cos@)

=> sin@+cos@-(sin@+cos@)sin@cos@=sin@+cos@+sin@cos@

Cancelling like terms and transposing to RHS

=> 0 = (sin@cos@)(1+sin@+cos@)

So either sin@=0, cos @=0 or sin@+cos@= -1