1. **State the problem:** Simplify the expression $$\left(\frac{3}{2}a + b\right)^3 - \frac{3}{2}a \left[\left(\frac{3}{2}a + b\right)^2 + (a - \frac{1}{4}b)(a + \frac{1}{4}b) - (a - \frac{3}{4}b)^2 + \frac{7}{3}b \left(2a + \frac{9}{8}b\right)\right].$$ 2. **Recall formulas and rules:** - Cube of a binomial: $$ (x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3 $$ - Square of a binomial: $$ (x + y)^2 = x^2 + 2xy + y^2 $$ - Difference of squares: $$ (x - y)(x + y) = x^2 - y^2 $$ - Expand and simplify step-by-step. 3. **Expand** $$\left(\frac{3}{2}a + b\right)^3$$: $$\left(\frac{3}{2}a\right)^3 + 3 \left(\frac{3}{2}a\right)^2 b + 3 \left(\frac{3}{2}a\right) b^2 + b^3 = \frac{27}{8}a^3 + \frac{27}{4}a^2 b + \frac{9}{2}a b^2 + b^3.$$ 4. **Expand** $$\left(\frac{3}{2}a + b\right)^2$$: $$\left(\frac{3}{2}a\right)^2 + 2 \cdot \frac{3}{2}a \cdot b + b^2 = \frac{9}{4}a^2 + 3 a b + b^2.$$ 5. **Simplify** $$(a - \frac{1}{4}b)(a + \frac{1}{4}b)$$ using difference of squares: $$a^2 - \left(\frac{1}{4}b\right)^2 = a^2 - \frac{1}{16}b^2.$$ 6. **Expand** $$(a - \frac{3}{4}b)^2$$: $$a^2 - 2 \cdot a \cdot \frac{3}{4}b + \left(\frac{3}{4}b\right)^2 = a^2 - \frac{3}{2} a b + \frac{9}{16} b^2.$$ 7. **Expand** $$\frac{7}{3} b \left(2 a + \frac{9}{8} b\right)$$: $$\frac{7}{3} b \cdot 2 a + \frac{7}{3} b \cdot \frac{9}{8} b = \frac{14}{3} a b + \frac{63}{24} b^2 = \frac{14}{3} a b + \frac{21}{8} b^2.$$ 8. **Sum inside the bracket:** $$\left(\frac{9}{4} a^2 + 3 a b + b^2\right) + \left(a^2 - \frac{1}{16} b^2\right) - \left(a^2 - \frac{3}{2} a b + \frac{9}{16} b^2\right) + \left(\frac{14}{3} a b + \frac{21}{8} b^2\right).$$ 9. **Combine like terms carefully:** - For $a^2$ terms: $$\frac{9}{4} a^2 + a^2 - a^2 = \frac{9}{4} a^2.$$ - For $a b$ terms: $$3 a b + 0 + \frac{3}{2} a b + \frac{14}{3} a b = 3 a b + \frac{3}{2} a b + \frac{14}{3} a b.$$ - For $b^2$ terms: $$b^2 - \frac{1}{16} b^2 - \frac{9}{16} b^2 + \frac{21}{8} b^2 = b^2 - \frac{10}{16} b^2 + \frac{21}{8} b^2 = b^2 - \frac{5}{8} b^2 + \frac{21}{8} b^2 = \left(1 - \frac{5}{8} + \frac{21}{8}\right) b^2 = \frac{24}{8} b^2 = 3 b^2.$$ 10. **Sum $a b$ coefficients:** $$3 + \frac{3}{2} + \frac{14}{3} = \frac{18}{6} + \frac{9}{6} + \frac{28}{6} = \frac{55}{6}.$$ So the bracket simplifies to: $$\frac{9}{4} a^2 + \frac{55}{6} a b + 3 b^2.$$ 11. **Multiply the bracket by** $$\frac{3}{2} a$$: $$\frac{3}{2} a \left(\frac{9}{4} a^2 + \frac{55}{6} a b + 3 b^2\right) = \frac{3}{2} a \cdot \frac{9}{4} a^2 + \frac{3}{2} a \cdot \frac{55}{6} a b + \frac{3}{2} a \cdot 3 b^2 = \frac{27}{8} a^3 + \frac{55}{4} a^2 b + \frac{9}{2} a b^2.$$ 12. **Subtract this from the cube term:** $$\left(\frac{27}{8} a^3 + \frac{27}{4} a^2 b + \frac{9}{2} a b^2 + b^3\right) - \left(\frac{27}{8} a^3 + \frac{55}{4} a^2 b + \frac{9}{2} a b^2\right) = \left(\frac{27}{8} a^3 - \frac{27}{8} a^3\right) + \left(\frac{27}{4} a^2 b - \frac{55}{4} a^2 b\right) + \left(\frac{9}{2} a b^2 - \frac{9}{2} a b^2\right) + b^3 = -\frac{28}{4} a^2 b + b^3 = -7 a^2 b + b^3.$$ 13. **Final simplified expression:** $$\boxed{-7 a^2 b + b^3}.$$