4sin³ A = 3sin A - sin 3A
So sin³ A = ¾sin A - ¼sin 3A

sin³ θ
+ ⅓ sin³ 3θ
+ 1/3² sin³ (3²θ)
+ ...
+ 1/3ⁿ sin³ (3ⁿθ)

=

¾sin θ - ¼sin 3θ
+ ⅓( ¾sin 3θ - ¼sin 9θ )
+ 1/9 ( ¾sin 9θ - ¼sin 27θ )
+...
+ 1/3ⁿ-¹ ( ¾sin (3ⁿ-¹θ) - ¼sin (3(3ⁿ-¹θ)) )
+ 1/3ⁿ ( ¾sin (3ⁿθ) - ¼sin (3(3ⁿθ)) )

=

¾sin θ - ¼sin 3θ
+ ¼sin 3θ - 1/12 sin 9θ
+ 1/12 sin 9θ - 1/36 sin 27θ
+...
+ 1/4 (3ⁿ-²) sin (3ⁿ-¹θ) - 1/4(3ⁿ-¹) sin (3ⁿθ)
+ 1/4(3ⁿ-¹) sin (3ⁿθ) - 1/4(3ⁿ) sin (3ⁿ+¹ θ)


(Cancel the common terms , method of difference)

= ¾ sinθ - 1/4(3ⁿ) sin (3ⁿ+¹ θ)

Deduction part :

Firstly,

sin² θ cos θ
= ½(2sin θ cos θ) sin θ
= ½sin 2θ sin θ

(Recall from MF26 that sin 2A = 2sin A cos A)

= ½ sin ((3θ + θ)/2) sin ((3θ - θ)/2)

= ½(-½) (-2sin ((3θ + θ)/2) sin ((3θ - θ)/2) )

= -¼ (cos 3θ - cos θ)

(Recall from MF26 that cos P + cos Q = -2sin ½(P+Q) sin ½(P - Q) )

= ¼ cos θ - ¼ cos 3θ


So, sin² θ cos θ + sin² 3θ cos 3θ + ... + sin² (3ⁿθ) cos² (3ⁿθ)

=

¼ cos θ - ¼ cos 3θ
+ ¼ cos 3θ - ¼ cos 3(3θ)
+ ¼ cos (3²θ) - ¼ cos 3(3²θ)
+ ...
+ ¼ cos (3ⁿ-¹ θ) - ¼ cos 3(3ⁿ-¹ θ)
+ ¼ cos (3ⁿ θ) - ¼ cos 3(3ⁿ θ)

=

¼ cos θ - ¼ cos 3θ
+ ¼ cos 3θ - ¼ cos 9θ
+ ¼ cos 9θ - ¼ cos 27θ
+ ...
+ ¼ cos (3ⁿ-¹ θ) - ¼ cos (3ⁿ θ)
+ ¼ cos (3ⁿ θ) - ¼ cos (3ⁿ+¹ θ)

(Cancel the common terms , method of difference)

= ¼ cos θ - ¼ cos (3ⁿ+¹ θ)

Alternatively,

4sin³ A = 3sin A - sin 3A
4sin³ A = 2sin A + sin A - sin 3A

4sin³ A = 2sin A + 2cos ((A + 3A)/2) sin ((A - 3A)/2)
(Recall factor formula from MF26)

4sin³ A = 2sin A + 2cos 2A sin (-A)
4sin³ A = 2sin A - 2cos 2A sin A

4sin² A = 2 - 2 cos 2A
4sin² A cos A = 2cos A - 2 cos 2A cos A

4sin² A cos A = 2cos A - 2 cos ((3A + A)/2) cos ((3A - A)/2)

4sin² A cos A = 2cos A - (cos 3A + cos A)
(Recall factor formula from MF26)

4 sin² A cos A = cos A - cos 3A

sin² A cos A = ¼ cos A - ¼ cos 3A

Solve accordingly.