AMC 10 · 2006 · #2
Grade 6 arithmeticDefine x⊗y=x3−y. What is h⊗(h⊗h)?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A new operation is defined by the rule $x \otimes y = x^3 - y$: cube the first number, then subtract the second. Using this rule, find $h \otimes (h \otimes h)$, where the same letter $h$ fills every slot.
Givens: The rule $x \otimes y = x^3 - y$; The expression to evaluate is $h \otimes (h \otimes h)$, with $h$ in every position; Answer choices: (A) $-h$, (B) $0$, (C) $h$, (D) $2h$, (E) $h^3$
Unknowns: The single expression in $h$ that $h \otimes (h \otimes h)$ equals
Understand
Restated: A new operation is defined by the rule $x \otimes y = x^3 - y$: cube the first number, then subtract the second. Using this rule, find $h \otimes (h \otimes h)$, where the same letter $h$ fills every slot.
Givens: The rule $x \otimes y = x^3 - y$; The expression to evaluate is $h \otimes (h \otimes h)$, with $h$ in every position; Answer choices: (A) $-h$, (B) $0$, (C) $h$, (D) $2h$, (E) $h^3$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities
The expression $h \otimes (h \otimes h)$ hides one operation inside another, so Tool #7 (Identify Subproblems) says: settle the parentheses first, then handle the outer $\otimes$. Tool #4 (Introduce a Variable) lets us hold the inner answer as one lump and drop it into the $y$ slot of the rule, so we never lose track of which input gets cubed. Tool #3 (Eliminate Possibilities) reads the choices — the outcome must stay a simple expression in $h$, and choice (E) $h^3$ is exactly the term that ends up cancelling, so it is a natural trap to watch for.
Execute — Answer: C
6.EE.A.2 Step 1 Do the inside operation first
- The parentheses come first, so compute $h \otimes h$.
- Feed it into the rule with $x = h$ and $y = h$: cube the first input and subtract the second, giving $h^3 - h$.
- That is the value the inner operation produces.
💡 A rule wrapped in parentheses has to be finished before anything outside it can use its result.
6.EE.A.2 Step 2 Feed that result into the outer rule
- Now the problem is $h \otimes (h^3 - h)$.
- Apply the rule again, this time with $x = h$ and $y = h^3 - h$.
- Only the first input is cubed, so the $x^3$ part is $h^3$, and the whole second input $h^3 - h$ gets subtracted as one block.
- Keeping it inside its own parentheses stops the subtraction sign from being applied to only part of it.
💡 The second input, however long, drops into the $y$ slot as a single quantity to be subtracted whole.
6.EE.A.3 Step 3 Distribute the subtraction and simplify
- Subtracting the block $(h^3 - h)$ means subtracting $h^3$ and adding back $h$: $h^3 - h^3 + h$.
- The two $h^3$ terms are opposites, so they cancel, leaving just $h$.
- That matches choice (C).
- Notice (E) $h^3$ is what you would land on if you forgot to subtract the inner $-h$, so the cancellation is the whole point of the problem.
💡 Subtracting a difference flips the sign of each piece inside, so the matching $h^3$ terms wipe each other out.
6.EE.A.2 The parentheses come first, so compute $h \otimes h$. Feed it into the rule with 6.EE.A.2 Now the problem is $h \otimes (h^3 - h)$. Apply the rule again, this time with $ 6.EE.A.3 Subtracting the block $(h^3 - h)$ means subtracting $h^3$ and adding back $h$: $ Review
Reasonableness: Test the rule on a number to be sure the letters were handled right. Let $h = 2$. Inner: $2 \otimes 2 = 2^3 - 2 = 6$. Outer: $2 \otimes 6 = 2^3 - 6 = 8 - 6 = 2$, which equals $h = 2$. The numeric result matches choice (C) and rules out the others: (A) $-h = -2$, (B) $0$, (D) $2h = 4$, and (E) $h^3 = 8$ are all wrong for $h = 2$, so only (C) survives.
Alternative: Work the algebra in one line without naming the inner value: $h \otimes (h \otimes h) = h \otimes (h^3 - h) = h^3 - (h^3 - h)$. Distributing the minus sign gives $h^3 - h^3 + h = h$, the same answer (C). The key is that the leading $h^3$ from the outer rule is exactly cancelled by the $-h^3$ hidden inside the second input.
CCSS standards used (min grade 6)
6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Applying the definition $x \otimes y = x^3 - y$ by substituting $h$ into the $x$ and $y$ slots, first for the inner operation and then for the outer one.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Reading and carrying the cube $h^3$ (and $2^3 = 8$ in the check) that the rule's $x^3$ term produces.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Distributing the subtraction in $h^3 - (h^3 - h)$ and cancelling the two $h^3$ terms to reach $h$.)
⭐ Finish the parentheses first, then plug that whole result into the rule as the part you subtract — here the two $h^3$ pieces cancel and only $h$ is left.
⭐ Finish the parentheses first, then plug that whole result into the rule as the part you subtract — here the two $h^3$ pieces cancel and only $h$ is left.
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