The coefficient of x cube in the expansion of (3+az)⁶ is 160 find the value of the constant a, hence find the coefficient of x cube in the expansion of (3+ ax)⁶(1-2x)

The general term in the expansion of $(A+B)^n$ is given by $T_{r+1} = \binom{n}{r} A^{n-r} B^r$.
For $(3+ax)^6$, we have $A=3$, $B=ax$, and $n=6$.
So, the general term is:

$$T_{r+1} = \binom{6}{r} (3)^{6-r} (ax)^r$$ $$T_{r+1} = \binom{6}{r} 3^{6-r} a^r x^r$$

For the term with $x^3$, we set $r=3$:

$$T_{3+1} = T_4 = \binom{6}{3} 3^{6-3} a^3 x^3$$ $$T_4 = \binom{6}{3} 3^3 a^3 x^3$$

$$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$ $$3^3 = 27$$

Substitute these values back into the expression for $T_4$:

$$T_4 = 20 \times 27 \times a^3 x^3$$ $$T_4 = 540 a^3 x^3$$

The coefficient of $x^3$ is $540a^3$.

We are given that the coefficient of $x^3$ is 160:

$$540 a^3 = 160$$ $$a^3 = \frac{160}{540}$$ $$a^3 = \frac{16}{54}$$ $$a^3 = \frac{8}{27}$$

Take the cube root of both sides:

$$a = \sqrt[3]{\frac{8}{27}}$$ $$a = \frac{2}{3}$$

The value of the constant $a$ is $\boxed{\frac{2}{3}}$.

We need the coefficients of $x^0$, $x^1$, $x^2$, and $x^3$ from $(3+\frac{2}{3}x)^6$.
Using the general term $T_{r+1} = \binom{6}{r} 3^{6-r} (\frac{2}{3}x)^r = \binom{6}{r} 3^{6-r} (\frac{2}{3})^r x^r$:

• Coefficient of $x^0$ (constant term, $r=0$): $$\binom{6}{0} 3^6 (\frac{2}{3})^0 = 1 \times 729 \times 1 = 729$$
• Coefficient of $x^1$ ($r=1$): $$\binom{6}{1} 3^5 (\frac{2}{3})^1 = 6 \times 243 \times \frac{2}{3} = 6 \times 81 \times 2 = 972$$
• Coefficient of $x^2$ ($r=2$): $$\binom{6}{2} 3^4 (\frac{2}{3})^2 = 15 \times 81 \times \frac{4}{9} = 15 \times 9 \times 4 = 540$$
• Coefficient of $x^3$ ($r=3$): $$\binom{6}{3} 3^3 (\frac{2}{3})^3 = 20 \times 27 \times \frac{8}{27} = 20 \times 8 = 160$$

So, $(3+\frac{2}{3}x)^6 = 729 + 972x + 540x^2 + 160x^3 + \dots$

We need the coefficient of $x^3$ in $(729 + 972x + 540x^2 + 160x^3 + \dots)(1-2x)$.
To get an $x^3$ term, we can multiply:

• The $x^3$ term from the first bracket by the constant term from the second bracket: $(160x^3) \times (1) = 160x^3$
• The $x^2$ term from the first bracket by the $x$ term from the second bracket: $(540x^2) \times (-2x) = -1080x^3$

The total coefficient of $x^3$ is the sum of these contributions:

$$160 + (-1080) = 160 - 1080 = -920$$

The coefficient of $x^3$ in the