Show that the equation $3e^{2x}+7e^{x}=6$ has only one solution and find its value correct to 2 significant figures. Factorise $(3x + 1)^3 - 8$ completely.

1. Express the equation as a quadratic in $e^x$.

Let $y = e^x$. The equation $3e^{2x} + 7e^x = 6$ becomes:
$3y^2 + 7y - 6 = 0$

2. Solve the quadratic equation for $e^x$.

$(3y - 2)(y + 3) = 0$
$y = \frac{2}{3} \quad \text{or} \quad y = -3$
$e^x = \frac{2}{3} \quad \text{or} \quad e^x = -3$

3. Show that there is only one valid solution.

Since $e^x > 0$ for all real $x$, $e^x = -3$ has no real solution.
Therefore, the only solution comes from $e^x = \frac{2}{3}$.

4. Find the value of $x$ correct to 2 significant figures.

$x = \ln\left(\frac{2}{3}\right)$
$x \approx -0.405465$
$x = -0.41 \text{ (to 2 s.f.)}$

1. Apply the difference of cubes identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Let $a = 3x + 1$ and $b = 2$.
$(3x + 1)^3 - 2^3 = [(3x + 1) - 2][(3x + 1)^2 + (3x + 1)(2) + 2^2]$

2. Simplify the factors

$= (3x - 1)(9x^2 + 6x + 1 + 6x + 2 + 4)$
$= (3x - 1)(9x^2 + 12x + 7)$