1. **State the problem:** Given the expansion of $$(ax - 2)^4 \left(1 + \frac{b}{x}\right)^3$$ written in descending powers of $x$, the first three terms are $$81x^4 + 999x^3 + cx^2$$. We need to find positive integers $a$, $b$, and $c$. 2. **Expand each part separately:** - Expand $$(ax - 2)^4$$ using the binomial theorem: $$ (ax - 2)^4 = \sum_{k=0}^4 \binom{4}{k} (ax)^k (-2)^{4-k} $$ Calculate each term: - For $k=4$: $\binom{4}{4} (ax)^4 (-2)^0 = 1 \cdot a^4 x^4 \cdot 1 = a^4 x^4$ - For $k=3$: $\binom{4}{3} (ax)^3 (-2)^1 = 4 a^3 x^3 (-2) = -8 a^3 x^3$ - For $k=2$: $\binom{4}{2} (ax)^2 (-2)^2 = 6 a^2 x^2 4 = 24 a^2 x^2$ - For $k=1$: $\binom{4}{1} (ax)^1 (-2)^3 = 4 a x (-8) = -32 a x$ - For $k=0$: $\binom{4}{0} (ax)^0 (-2)^4 = 1 \cdot 1 \cdot 16 = 16$ Hence, $$ (ax - 2)^4 = a^4 x^4 - 8 a^3 x^3 + 24 a^2 x^2 - 32 a x + 16 $$ 3. **Expand $$\left(1 + \frac{b}{x}\right)^3$$ via binomial theorem:** $$ \sum_{m=0}^3 \binom{3}{m} 1^{3-m} \left(\frac{b}{x}\right)^m $$ Calculate each term: - For $m=0$: $1$ - For $m=1$: $3 \frac{b}{x} = \frac{3b}{x}$ - For $m=2$: $3 \frac{b^2}{x^2} = \frac{3 b^2}{x^2}$ - For $m=3$: $\frac{b^3}{x^3}$ So, $$ \left(1 + \frac{b}{x}\right)^3 = 1 + \frac{3b}{x} + \frac{3 b^2}{x^2} + \frac{b^3}{x^3} $$ 4. **Multiply the expansions:** We want terms up to $$x^2$$ in the product: $$ (ax - 2)^4 \left(1 + \frac{b}{x}\right)^3 = \left(a^4 x^4 - 8 a^3 x^3 + 24 a^2 x^2 - 32 a x + 16 \right) \left(1 + \frac{3b}{x} + \frac{3 b^2}{x^2} + \frac{b^3}{x^3}\right) $$ Multiply term-by-term focusing only on terms with powers $x^4$, $x^3$, and $x^2$: - Terms giving $x^4$: - $a^4 x^4 \cdot 1 = a^4 x^4$ - Terms giving $x^3$: - $a^4 x^4 \cdot \frac{3b}{x} = 3 a^4 b x^3$ - $-8 a^3 x^3 \cdot 1 = -8 a^3 x^3$ Sum for $x^3$ term: $$ (3 a^4 b - 8 a^3) x^3 $$ - Terms giving $x^2$: - $a^4 x^4 \cdot \frac{3 b^2}{x^2} = 3 a^4 b^2 x^2$ - $-8 a^3 x^3 \cdot \frac{3b}{x} = -24 a^3 b x^2$ - $24 a^2 x^2 \cdot 1 = 24 a^2 x^2$ Sum for $x^2$ term: $$ (3 a^4 b^2 - 24 a^3 b + 24 a^2) x^2 $$ 5. **Match coefficients with given terms:** Given first three terms are: $$81 x^4 + 999 x^3 + c x^2$$ Equate coefficients: - For $x^4$: $$a^4 = 81$$ - For $x^3$: $$3 a^4 b - 8 a^3 = 999$$ - For $x^2$: $$3 a^4 b^2 - 24 a^3 b + 24 a^2 = c$$ 6. **Find $a$:** Since $a$ is a positive integer, $$ a^4 = 81 = 3^4 \implies a = 3 $$ 7. **Find $b$: substitute $a=3$ into the $x^3$ coefficient equation:** $$3 (3^4) b - 8 (3^3) = 999$$ $$3 \times 81 b - 8 \times 27 = 999$$ $$243 b - 216 = 999$$ $$243 b = 1215$$ $$b = \frac{1215}{243} = 5$$ 8. **Find $c$: substitute $a=3$, $b=5$ into $x^2$ coefficient:** $$c = 3 (3^4)(5^2) - 24 (3^3)(5) + 24 (3^2)$$ $$= 3 \times 81 \times 25 - 24 \times 27 \times 5 + 24 \times 9$$ $$= 3 \times 2025 - 3240 + 216$$ $$= 6075 - 3240 + 216 = 3051$$ **Final answers:** $$a=3, \quad b=5, \quad c=3051$$