Use the real cube root \(t=\sqrt[3]x\), so \(x^{2/3}=t^2\) and \((8x)^{1/3}=2t\). Then \[t^2+2t-24=(t+6)(t-4)=0.\] Hence \(t=-6\) or \(t=4\), giving \(\boxed{x=-216\ \text{or}\ x=64}\). Both satisfy the original real-root equation.

Since \(x\ne0\), multiply by \(x^2\): \[3x^4-8x^2-16=0.\] Put \(u=x^2\). Then \[3u^2-8u-16=(3u+4)(u-4)=0.\] Reject \(u=-4/3\), since \(u=x^2\ge0\). Thus \(x^2=4\), giving \(\boxed{x=\pm2}\).

(a) \(x\sqrt x=x^1x^{1/2}=\boxed{x^{3/2}}\).

(b) The real domain is \(x\ge0\). Put \(u=x^{3/2}\), so \(x^3=u^2\). Then \[u^2-9u+8=(u-1)(u-8)=0.\] Therefore \(x=1^{2/3}=1\) or \(x=8^{2/3}=4\): \(\boxed{x=1,4}\).

Let \(u=x^3\). Then \(8u^2+215u-27=(8u-1)(u+27)=0\). Thus \(x^3=1/8\) or \(-27\), giving the real solutions \(\boxed{x=1/2,-3}\).