=1(1+sin‌θ‌cos‌θ) =1+sin‌θ‌cos‌θ.....(i) [∵sin‌θ−cos‌θ=1] On squaring both sides, (sin‌θ−cos‌θ)2=(1)2 ⇒(sin2θ+cos2θ−2‌sin‌θ‌cos‌θ)=1 ⇒‌‌1−2‌sin‌θ‌cos‌θ=1 ⇒‌‌sin‌θ‌cos‌θ=0.....(ii) From Eqs. (ii) and (i), we get sin3θ−cos3θ=1