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?

In an equivalent-expression question with two letters, choose allowed perfect cubes for the letters and compare the original with each choice in Desmos. The cube roots then become whole numbers. A different value rules a choice out; one matching value alone does not prove equivalence. If several choices match, test another pair or use factoring to settle them.

Hints

Hint 1
An equivalent expression has the same value as the original for every allowed pair. Choose two unequal positive perfect cubes so the cube roots are quick to evaluate. What happens if you use the same number for both letters?
Hint 2
Evaluate the original first. Then evaluate each choice using the same values of $a$ and $b$ . A choice with a different result cannot be equivalent, even if it looks similar to the original.

Step-by-step

Approach 1: Test the choices with perfect cubes

Step 1
Choose values that make the roots manageable
Choose $a=8$ and $b=1$ . They are positive, unequal perfect cubes: $\sqrt[3]{8}=2$ and $\sqrt[3]{1}=1$ . Unequal matters because $a=b$ would make denominators in the original expression zero.
Step 2
Evaluate the original expression
Type $a=8$ and $b=1$ , then type the entire original expression on the next line. Desmos displays $3$ . Any equivalent choice must also give $3$ for this allowed pair.
Step 3
Compare all four choices
Type each choice on its own line. Desmos displays $3$ , $7$ , $5$ , and $1$ , in the order given. Only $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$ matches the original's $3$ . One input can rule out a choice, even though one match alone cannot prove an identity. Here, the other three choices are ruled out. Choice A.

Approach 2: See why the match works for every allowed pair

Step 1
Factor the first numerator
The first numerator is a difference of squares. Factor it, keeping both factors: $a^2-b^2=(a-b)(a+b)$ .
Step 2
Cancel the matching factor
Since $a\ne b$ , the factor $a-b$ is not zero, so it can cancel across the product: $\frac{(a-b)(a+b)}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\cdot\frac{\sqrt[3]{a}-\sqrt[3]{b}}{a-b}=\frac{(a+b)(\sqrt[3]{a}-\sqrt[3]{b})}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}$ . Cancel whole factors, not individual terms inside a difference.
Step 3
Name the cube roots
Let $u=\sqrt[3]{a}$ and $v=\sqrt[3]{b}$ . Then $a=u^3$ , $b=v^3$ , and squaring each cube root gives $\sqrt[3]{a^2}=u^2$ and $\sqrt[3]{b^2}=v^2$ . Rewrite the fraction: $\frac{(a+b)(\sqrt[3]{a}-\sqrt[3]{b})}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}=\frac{(a+b)(u-v)}{u^2-v^2}$ .
Step 4
Factor the new denominator
Use the difference of squares again. Factor the denominator: $\frac{(a+b)(u-v)}{u^2-v^2}=\frac{(a+b)(u-v)}{(u-v)(u+v)}$ . Because $a\ne b$ , their cube roots differ, so $u-v$ is nonzero. Cancel it: $\frac{(a+b)(u-v)}{(u-v)(u+v)}=\frac{a+b}{u+v}$ .
Step 5
Factor the sum of cubes
From $a=u^3$ and $b=v^3$ , the numerator is a sum of cubes. Its middle term has a minus sign so the cross terms cancel when you multiply the factors. Rewrite and factor: $\frac{a+b}{u+v}=\frac{u^3+v^3}{u+v}=\frac{(u+v)(u^2-uv+v^2)}{u+v}$ . Since $u$ and $v$ are positive, cancel $u+v$ : $\frac{(u+v)(u^2-uv+v^2)}{u+v}=u^2-uv+v^2$ .
Step 6
Return to the answer's notation
Replace $u$ and $v$ with their cube roots. The product $uv$ is $\sqrt[3]{ab}$ , so the equivalent expression is $u^2-uv+v^2=\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$ . Choice A.

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

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