Question: $P$ varies directly with the cube root of $x$, and $P = 2$ when $x = 27$. Express $P$ in terms of $x$, and find $x$ when $P = \dfrac{4}{3}$.

Step 1. Translate the sentence into an equation with a constant $k$:

Step 2. Find $k$ from the given pair $(x, P) = (27, 2)$:

Step 3. Use the equation for the new value:

Step 4. Undo the cube root by cubing:

Check. $P = \frac{2}{3}\sqrt[3]{8} = \frac{2}{3}(2) = \frac{4}{3}$. ✓

Graph check: sqrt[3]{x} a straight line through the origin — the signature of every direct variation.

⚠ Watch out: "Directly proportional to the cube root of $x$" is $P = k\sqrt[3]{x}$ — not $P = k x^3$ (that's the cube) and not $P = \sqrt[3]{kx}$. Always find $k$ first from the given data point; the equation with $k$ evaluated is what every later part runs on.

Question: $y$ is inversely proportional to the cube of $x$, and $y = 5$ when $x = 2$. Find the equation relating $y$ and $x$, the value of $y$ when $x = 4$, and the value of $x$ when $y = 10$.

Step 1. Inverse proportion puts the power of $x$ underneath:

Step 2. Find $k$ from $(x, y) = (2, 5)$:

Step 3. When $x = 4$:

(Doubling $x$ divided $y$ by $2^3 = 8$ — exactly what "inverse cube" means.)

Step 4. When $y = 10$:

Graph check: dfrac{1}{x^3} a straight line through the origin, and as — never a line with a negative gradient.

⚠ Watch out: Inverse means $\dfrac{k}{x^3}$, never $y = k - x^3$ or $-kx^3$. In the last step the unknown is trapped under the fraction — multiply up first ($10x^3 = 40$), then take the cube root, not the square root.