Question 5·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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For positive values of $a$ and $b$ , where $a\ne b$ , the expression

\begin{aligned} &\left(\frac{a^2-b^2}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\right) \\[1.4em] &\qquad\cdot \left(\frac{\sqrt[3]{a}-\sqrt[3]{b}}{a-b}\right) \end{aligned}

is equivalent to which choice?

For equivalent expressions in two variables, a numerical test can eliminate complicated choices quickly. Choose positive, unequal perfect cubes, then evaluate the original and each choice at the same pair in Desmos. One mismatch rules a choice out; one match alone doesn't prove equivalence. If you need the exact reason, name the cube roots as new variables and factor.

Hints

Hint 1
A perfect cube has a whole-number cube root. Try the positive, unequal values $8$ and $1$ : their cube roots are $2$ and $1$ . What does the original product give at that pair?
Hint 2
Equivalent expressions must agree at every allowed pair. A single mismatch eliminates a choice, so compare each choice with the original using the same values of $a$ and $b$ .

Step-by-step

Approach 1: Eliminate choices with Desmos

Step 1
Choose a pair with clean cube roots
Pick the positive, unequal pair $a=8$ and $b=1$ . Both are perfect cubes, so their cube roots are whole numbers. Type $a=8$ and $b=1$ on separate Desmos lines; Desmos sets those values for the expressions below.
Step 2
Find the original expression's output
Type the whole product, not one fraction: $\frac{a^2-b^2}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\cdot\frac{\sqrt[3]{a}-\sqrt[3]{b}}{a-b}$ . Desmos prints $3$ , so an equivalent choice must also give $3$ at this pair.
Step 3
Rule out three choices
Type $\sqrt[3]{a^2}+\sqrt[3]{ab}+\sqrt[3]{b^2}$ , $\sqrt[3]{a^2}+\sqrt[3]{b^2}$ , and $\sqrt[3]{a}-\sqrt[3]{b}$ on new lines. Desmos prints $7$ , $5$ , and $1$ . These mismatches with $3$ rule out all three expressions.
Step 4
Check the remaining expression
Type $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$ last. Desmos prints $3$ , matching the original. A single match alone wouldn't prove equivalence, but the other three choices failed at an allowed pair. So this remaining expression is the equivalent one. Choice A.

Approach 2: See why the match is exact

Step 1
Replace cube roots with new variables
Let $u=\sqrt[3]{a}$ and $v=\sqrt[3]{b}$ , and call the original product $E$ . Then $a=u^3$ , $b=v^3$ , and $\sqrt[3]{a^2}=u^2$ . Replace each cube-root term the same way:
$E=\frac{u^6-v^6}{u^2-v^2}\cdot\frac{u-v}{u^3-v^3}.$
Step 2
Expose matching factors
Use the difference of squares rule, $X^2-Y^2=(X-Y)(X+Y)$ , on both $u^6-v^6$ and $u^2-v^2$ :
$E=\frac{(u^3-v^3)(u^3+v^3)}{(u-v)(u+v)}\cdot\frac{u-v}{u^3-v^3}.$
Step 3
Cancel whole factors
The given $a\ne b$ means $u\ne v$ , so neither $u-v$ nor $u^3-v^3$ is zero. Cancel those matching factors:
$E=\frac{u^3+v^3}{u+v}.$
Step 4
Factor the sum of cubes
Use the sum of cubes pattern. The minus sign in the middle makes the cross terms cancel when $(u+v)(u^2-uv+v^2)$ is multiplied out. Replace the numerator with that product:
$E=\frac{(u+v)(u^2-uv+v^2)}{u+v}.$
Step 5
Cancel the last common factor
Both $u$ and $v$ are positive, so $u+v$ isn't zero. Cancel it:
$E=u^2-uv+v^2.$
Step 6
Return to the form in the choices
Put $u=\sqrt[3]{a}$ and $v=\sqrt[3]{b}$ back in. In particular, the middle term is $uv=\sqrt[3]{ab}$ : $E=\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$ . This is the expression equivalent to the original product. Choice A.

Question 5·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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