Factorise $\left(2x+1\right)^3 - \left(2x-1\right)^3$ completely. Hence show that the equation $\left(2x+1\right)\bigl[\left(2x+1\right)^3 - \left(2x-1\right)^3\bigr]=0$ has only one real root.

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

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

2. Simplify the resulting expression to find the completely factorised form

$= [2][(4x^2 + 4x + 1) + (4x^2 - 1) + (4x^2 - 4x + 1)]$
$= 2(12x^2 + 1)$

1. Substitute the result from part (a) into the given equation

From part (a), we have $(2x+1)^3 - (2x-1)^3 = 2(12x^2+1)$.
Substituting this into the equation:
$(2x+1)[2(12x^2+1)] = 0$
$2(2x+1)(12x^2+1) = 0$

2. Solve the linear factor for a real root

Setting the first factor to zero:
$2x+1 = 0$
$x = -\frac{1}{2}$

3. Show that the quadratic factor yields no real roots

Setting the second factor to zero:
$12x^2+1 = 0$
$x^2 = -\frac{1}{12}$
Since the square of any real number cannot be negative ($x^2 \ge 0$ for all real $x$), this equation has no real roots.
Alternatively, using the discriminant for $12x^2+0x+1=0$:
$b^2 - 4ac = 0^2 - 4(12)(1) = -48 < 0$
Since the discriminant is negative, there are no real roots from this factor.

4. Conclude the proof

Since the only real solution comes from the linear factor, the equation has only one real root, $x = -\frac{1}{2}$.