Super-Hard SAT Advanced Math Question 5 | Free SAT Prep

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

is equivalent to which choice?

For expressions involving several cube roots, replace each original variable with a cube, such as $a=x^3$ and $b=y^3$ . This often turns radical expressions into ordinary powers, making difference-of-squares and difference-of-cubes factoring patterns visible. Factor before multiplying everything out, then cancel only factors that are known to be nonzero from the given restrictions.

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Use a substitution

Set $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$ . Rewrite every occurrence of $a$ and $b$ in terms of $x$ and $y$ .

Look for differences of powers

After the substitution, identify expressions of the forms $x^6-y^6$ , $x^3-y^3$ , and $x^2-y^2$ .

Factor the final numerator

Once you combine the simplified factors, try factoring $x^4+x^2y^2+y^4$ as a product of two quadratic expressions.

Desmos Guide

Prefer algebra for an exact result

Factoring is faster here because the question asks for an expression equivalent for all allowed values. Desmos can be used to check the result with a convenient pair of values.

Choose values and evaluate the original expression

Enter a=27 and b=8. Then enter

E=((a^2-b^2)/(a^(2/3)-b^(2/3)))*((a^(1/3)-b^(1/3))/(a-b))

Desmos evaluates $E$ as $7$ .

Check the proposed simplified expression

Enter

A=a^(2/3)-(a*b)^(1/3)+b^(2/3)

This also evaluates to $7$ , supporting the algebraic simplification.

Step-by-step Explanation

Rewrite using cube-root variables

Let $x=\sqrt[3]{a}$ and $y=\sqrt[3]{b}$ . Then $a=x^3$ and $b=y^3$ , so the expression becomes

\left(\frac{x^6-y^6}{x^2-y^2}\right) \left(\frac{x-y}{x^3-y^3}\right).

Because $a\ne b$ , it follows that $x\ne y$ , so the denominators are nonzero.

Factor each difference

Use differences of powers:

\frac{x^6-y^6}{x^2-y^2}=x^4+x^2y^2+y^4

and

\frac{x-y}{x^3-y^3}=\frac{1}{x^2+xy+y^2}.

Therefore, the product is

\frac{x^4+x^2y^2+y^4}{x^2+xy+y^2}.

Factor and simplify the remaining quotient

The numerator factors as

x^4+x^2y^2+y^4=(x^2+xy+y^2)(x^2-xy+y^2).

Canceling the common factor gives

x^2-xy+y^2.

Substitute back

Since $x^2=\sqrt[3]{a^2}$ , $xy=\sqrt[3]{ab}$ , and $y^2=\sqrt[3]{b^2}$ , the result is

\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation