Answer: (a) no real roots; (b) \(x=8\) or \(x=-216\).

We start with the main method. Use the quadratic formula exactly and simplify the surds by rationalising where necessary.

For part (a), multiply both sides by \(5x-3\):

\((4x+1)(3x+2)=(x+1)(5x-3).\)

Expanding gives

\(12x^2+11x+2=5x^2+2x-3.\)

So

\(7x^2+9x+5=0.\)

The discriminant is

\(\Delta=9^2-4(7)(5)=81-140=-59.\)

Since \(\Delta\lt 0\), the equation has no real roots.

For part (b), let \(y=\sqrt[3]{x}\). Then

\(\frac{12}{y}-y=4.\)

Multiplying by \(y\),

\(12-y^2=4y.\)

Hence

\(y^2+4y-12=0.\)

Factorising,

\((y+6)(y-2)=0.\)

So \(y=2\) or \(y=-6\). Therefore

\(\sqrt[3]{x}=2\) or \(\sqrt[3]{x}=-6\), giving

\(x=8\) or \(x=-216.\)

This completes the solution and gives the required result.