A Cubic Identity

Alignments to Content Standards: A-SSE.B.3.a A-SSE.A.2

Task

Let $a$ and $b$ be real numbers with $a>b>0$ and $\frac{a^3-b^3}{(a-b)^3}=\frac{73}{3}$ . What is $\frac{b}{a}$ ?

IM Commentary

This task presents a challenging exercise in both algebraic manipulations and seeing structure in algebraic expressions (MP 7). In the solutions provided, students have to identify that $(a-b)$ is a factor of $(a^3-b^3)$ in order to simplify the expressions appearing to quadratics. At that point, either factoring or the quadratic formula can be employed to compute $\frac{b}{a}$ .

This task was adapted from problem #17 on the 2012 American Mathematics Competition (AMC) 10A Test, which asked students for the value $a-b$ given that $a$ and $b$ were relatively prime integers. (In the end, this forces $a=10$ and $b=7$ , so $a-b=3$ .) For the 2012 AMC 10A, which was taken by 73,703 students, the multiple choice answers for the problem had the following distribution:

Choice	Answer	Percentage of Answers
(A)	1	6.15
(B)	2	5.11
(C)	3	25.76
(D)	4	5.08
(E)	5	2.44
Omit	--	55.42

Of the 73,703 students: 36206, or 49%, were in 10th grade;25,498 or 35%, were in 9th grade; and the remainder were below than 9th grade.

Solutions

Solution: 1 Algebra, then Factoring

Vital to solving this problem is the fact that the numerator is a difference of cubes, which suggests using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ . Using this identity, and that $a\neq b$ , we can simplify the expression

$\begin{align} \frac{a^3 - b^3}{(a - b)^3} &= \frac{(a-b)(a^2 + ab + b^2)}{(a-b)(a^2 -2ab + b^2) } \\ &= \frac{ a^2 + ab + b^2 }{ a^2 - 2ab + b^2 }. \end{align}$

The equation $\frac{a^3-b^3}{(a-b)^3} = \frac{73}{3}$ can be written, using the identity from the previous paragraph, as as

$\begin{align} 3(a^2 + ab +b^2) = 73(a^2 - 2ab + b^2). \end{align}$

Gathering all terms on the same side, this is the same as

$70a^2-149ab+70b^2 = 0$ . Fortunately, this quadratic can be factored

$\begin{align} 70a^2-149ab+70b^2 = (10a - 7b)(7a -10b). \end{align}$

Because $a \gt b \gt 0$ , we are looking for a solution to $7a - 10b = 0$ rather than $10a - 7b = 0$ , which gives $7a=10b$ . Re-writing, we find $\frac{b}{a}=\frac{7}{10}=0.7.$

Solution: 2. Algebra, then Quadratic Formula

This solution is identical to the first until the quadratic relation between $a$ and $b$ is reached, at which point we instead use here the quadratic formula.

Vital to solving this problem is the fact that the numerator is a difference of cubes, which suggests using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ .

Using this identity, and that $a\neq b$ , we can simplify the expression

$\begin{align} \frac{a^3 - b^3}{(a - b)^3} &= \frac{(a-b)(a^2 + ab + b^2)}{(a-b)(a^2 -2ab + b^2) } \\ &= \frac{ a^2 + ab + b^2 }{ a^2 - 2ab + b^2 }. \end{align}$

The equation $\frac{a^3-b^3}{(a-b)^3} = \frac{73}{3}$ can be written, using the identity from the previous paragraph, as as

$\begin{align} 3(a^2 + ab +b^2) = 73(a^2 - 2ab + b^2). \end{align}$

Gathering all terms on the same side, this is the same as $70a^2-149ab+70b^2 = 0$ . Dividing both sides by $a^2$ , we are left with the quadratic equation $70\left(\frac{b}{a}\right)^2-149\left(\frac{b}{a}\right)+70=0.$ Using the quadratic formula gives $\frac{b}{a}=\frac{149\pm\sqrt{149^2-4\cdot 70^2}}{140}=\frac{149\pm 51}{40}=\frac{10}{7}\text{ or }\frac{7}{10}.$ Since $a \gt b \gt 0$ , $\frac{7}{10}$ must be the solution we are looking for.

Alternatively, we could apply the quadratic formula directly to the $(70)a^2-(149b)a+(70b^2)=0$ to get $a=\frac{149b\pm \sqrt{(149b)^2-4\cdot 70\cdot (70b^2)}}{140}=\frac{(149\pm 51)b}{140},$ giving to the same solution.