(i) Given that $$ 81 x^{4}-6 x^{3}-9 a^{2} x^{2}-11 x-3 \frac{8}{9} $$ is divisible by $$ 3 x+a $$, show that $$ 2 a^{3}+33 a-35=0 $$. ii) Solve the equation $$ 2 a^{3}+33 a-35=0 $$ for all real values of $$ a $$.

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**(i)** Since $$3x + a$$ is a factor of 
$$
81x^4 - 6x^3 - 9a^2x^2 - 11x - 3\frac{8}{9},
$$ 
its root
$$
3x + a = 0 \quad \Longrightarrow \quad x = -\frac{a}{3}
$$
must make the polynomial equal to zero. Note that 
$$
3\frac{8}{9} = 3 + \frac{8}{9} = \frac{27}{9} + \frac{8}{9} = \frac{35}{9}.
$$
Thus the polynomial can be written as
$$
81x^4 - 6x^3 - 9a^2x^2 - 11x - \frac{35}{9}.
$$

Substitute $$x = -\frac{a}{3}$$ into the polynomial:

1. The first term:
   $$
   81\left(-\frac{a}{3}\right)^4 = 81 \cdot \frac{a^4}{81} = a^4.
   $$
2. The second term:
   $$
   -6\left(-\frac{a}{3}\right)^3 = -6\left(-\frac{a^3}{27}\right) = \frac{6a^3}{27} = \frac{2a^3}{9}.
   $$
3. The third term:
   $$
   -9a^2\left(-\frac{a}{3}\right)^2 = -9a^2\cdot \frac{a^2}{9} = -a^4.
   $$
4. The fourth term:
   $$
   -11\left(-\frac{a}{3}\right) = \frac{11a}{3}.
   $$
5. The constant term remains:
   $$
   -\frac{35}{9}.
   $$

Summing these results, we have:
$$
a^4 + \frac{2a^3}{9} - a^4 + \frac{11a}{3} - \frac{35}{9} = 0.
$$
The $$a^4$$ and $$-a^4$$ cancel, leaving:
$$
\frac{2a^3}{9} + \frac{11a}{3} - \frac{35}{9} = 0.
$$
Multiply the entire equation by 9 to eliminate fractions:
$$
2a^3 + 33a - 35 = 0.
$$
Thus, we have shown that 
$$
2a^3 + 33a - 35 = 0.
$$

---

**(ii)** To solve 
$$
2a^3 + 33a - 35 = 0,
$$
we first look for rational roots. Testing $$a = 1$$:
$$
2(1)^3 + 33(1) - 35 = 2 + 33 - 35 = 0,
$$
so $$a = 1$$ is a solution.

Now, factor $$2a^3 + 33a - 35$$ by dividing it by $$(a - 1)$$. Write the polynomial in standard form with all degrees:
$$
2a^3 + 0a^2 + 33a - 35.
$$

Using synthetic division with root $$a = 1$$:

- Bring down the coefficient $$2$$.
- Multiply by $$1$$: $$2$$, add to $$0$$ to get $$2$$.
- Multiply $$2$$ by $$1$$: $$2$$, add to $$33$$ to get $$35$$.
- Multiply $$35$$ by $$1$$: $$35$$, add to $$-35$$ to get $$0$$.

Thus, the polynomial factors as:
$$
2a^3 + 33a - 35 = (a - 1)(2a^2 + 2a + 35).
$$

Next, solve the quadratic factor:
$$
2a^2 + 2a + 35 = 0.
$$
Compute the discriminant:
$$
\Delta = (2)^2 - 4(2)(35) = 4 - 280 = -276.
$$
Since the discriminant is negative, the quadratic has no real solutions.

Thus, the only real solution is:
$$
a = 1.
$$
$$ a = 1 $$

(i) Given that is
divisible by , show that .
ii) Solve the equation for all
real values of .

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