Verified Answer :
\(\bf \cfrac{cos^3(\theta)+sin^3(\theta)}{sin(\theta)cos(\theta)+sin^2(\theta)}=csc(\theta)-cos(\theta)\\\\
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\textit{let us do the left-hand-side}
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recall\qquad \textit{difference of cubes}
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a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\\\\)
\(\bf -----------------------------\\\\
thus
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\cfrac{[cos(\theta)+sin(\theta)][cos^2(\theta)-cos(\theta)sin(\theta)+sin^2(\theta)]}{sin(\theta)cos(\theta)+sin^2(\theta)}
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\textit{let us take common factor on the denominator}
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\cfrac{[\boxed{cos(\theta)+sin(\theta)}][cos^2(\theta)-cos(\theta)sin(\theta)+sin^2(\theta)]}{sin(\theta)[\boxed{cos(\theta)+sin(\theta)}]}
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\cfrac{cos^2(\theta)-cos(\theta)sin(\theta)+sin^2(\theta)}{sin(\theta)}\\\\
-----------------------------\\\\\)
\(\bf \textit{now, recall your pythagorean identities}\qquad sin^2(\theta)+cos^2(\theta)=1\\\\
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\cfrac{\boxed{cos^2(\theta)+sin^2(\theta)}-cos(\theta)sin(\theta)}{sin(\theta)}\implies
\cfrac{1-cos(\theta)sin(\theta)}{sin(\theta)}
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\textit{now, distributing the denominator}
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\cfrac{1}{sin(\theta)}-\cfrac{cos(\theta)\boxed{sin(\theta)}}{\boxed{sin(\theta)}}\implies \implies csc(\theta)-cos(\theta)\)
(cos^3Ф+sin^3Ф)/sinФcosФ+sin^2Ф=cosecФ-cosФ
lets take left hand side first
(cos^3Ф+sin^3Ф)/sinФcosФ+sin^2Ф
(A^3+B^3)=(A+B)(A^2-AB+B^2)
(cosФ+sinФ)(cos^2Ф-cosФsinФ+sin^2Ф)/sinФcosФ+sin^2Ф
(sin^2x+cos^2x=1)
(cosФ+sinФ)(1+sinФcos)/sinФcosФ+sin^2Ф
(cosФ+sinФ)1/2(2-2sinФcosФ) /sinФcosФ+sin^2Ф
1/2(cosФ+sin)(2-sin2Ф)/sinФcosФ+sin^2Ф
1/2(2-sin2Ф)/sinФ
now lets take RHS
cosecФ-cosФ=1/sinФ-cosФ
(1-sinФcosФ)/sinФ
1/2(2-2sinФcosФ)/sinФ
1/2*(2-2sin2Ф)/sinФ
Hence proved