For which of the following values of $x$ is $x^{3}<x^{2}$ ?
(A) $x=\frac{5}{3}$
(B) $x=\frac{3}{4}$
(C) $x=1$
(D) $x=\frac{3}{2}$
(E) $x=\frac{21}{20}$
If $x=1$, then $x^{2}=1$ and $x^{3}=1$ and so $x^{3}=x^{2}$.

If $x>1$, then $x^{3}$ equals $x$ times $x^{2}$; since $x>1$, then $x$ times $x^{2}$ is greater than $x^{2}$ and so $x^{3}>x^{2}$.

Therefore, if $x$ is positive with $x^{3}<x^{2}$, we must have $0<x<1$. We note that if $0<x<1$, then both $x, x^{2}$ and $x^{3}$ are all positive, and $x^{3}=x^{2} \times x<x^{2} \times 1=x^{2}$.

Of the given choices, only $x=\frac{3}{4}$ satisfies $0<x<1$, and so the answer is (B).

We can verify by direct calculation that this is the only correct answer from the given choices.

ANSWER: (B)