Let $t = x^{1/3}$. What does $x^{2/3}$ become in terms of $t$, and what kind of equation in $t$ do you get?

Once you have a quadratic in $t$, factor it to find the values of $t$. Then remember $t = x^{1/3}$—what must you do to get $x$?

You should get two real values for $x$. Make sure you identify which one is larger to answer the question.

Type f(x) = x^(2/3) - 5x^(1/3) + 6 into Desmos. Make sure you use parentheses like x^(1/3) for the cube root power.

Look at the graph of $f(x)$ and identify where it crosses the x-axis (where $f(x) = 0$). You can tap on the x-intercepts or add a table to see the exact x-values of the roots.

You should see two x-intercepts. Compare their x-values and note which one is larger; that is the greater real solution asked for in the question.

Notice that the exponents $2/3$ and $1/3$ are related: $x^{2/3} = (x^{1/3})^2$. Let $t = x^{1/3}$. Then $x^{2/3} = t^2$.

Substitute into the equation:

Now you have a quadratic equation in $t$.

Factor the quadratic $t^2 - 5t + 6$:

So the solutions for $t$ are

Recall that $t = x^{1/3}$, so

Cube both sides of each equation to solve for $x$:

• If $x^{1/3} = 2$, then $x = 2^3 = 8$.
• If $x^{1/3} = 3$, then $x = 3^3 = 27$.

So the real solutions are $x = 8$ and $x = 27$.

The two real solutions are $8$ and $27$.

The greater real solution is $x = 27$, which corresponds to choice C.