Category: Blog

A condition connecting variables can simplify an expression before any individual variable is known. The key is to use an identity with a valid derivation, rather than assume that taking a power preserves addition.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send AMC12 CUBIC IDENTITY 1029 and tell us your school year and question.

State and verify the factorisation

The relevant identity is a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca).

Expanding the right-hand side cancels the mixed squared terms and leaves the stated cubic expression. This algebraic identity is valid before any zero-sum condition is imposed.

Apply the condition exactly where it belongs

If a + b + c = 0, the first factor on the right vanishes. Consequently a³ + b³ + c³ = 3abc.

For the original numerical example a = 2, b = 3 and c = -5, the cube sum is 8 + 27 - 125 = -90. The product expression gives 3 × 2 × 3 × (-5) = -90 as well.

Do not cube a sum term by term

Although (a + b + c)³ is zero under the condition, it does not follow that a³ + b³ + c³ is zero. Expanding a cube of a sum produces cross terms.

The numerical example is a direct counterexample to that tempting shortcut: its three variables sum to zero, but their cubes sum to minus ninety.

Know what the relation can and cannot determine

If the sum is zero and abc is known, the identity determines the cube sum immediately. If only the zero sum is known, the cube sum need not have one fixed numerical value because the product can vary.

This is an original AMC 12 learning exercise, not an official problem. A strong review includes both the factorisation and a counterexample to the incorrect rule, so the shortcut remains attached to its precise condition.

Continue reading: AMC 12 Sequences: Derive a Geometric Sum by Subtracting Two Lines.

Contact Hanlin

Planning a first attempt or reviewing a topic you find difficult? Scan the QR code and send AMC12 CUBIC IDENTITY 1029. Include your school year, country or competition region, experience and the question you want to discuss.

Contact Hanlin on WeChat

On your phone, use the button or save the QR image and scan it within WeChat.

Official information and further reading: Source 1

When a polynomial is divided by a quadratic, the remainder can be linear. Evaluating at roots of the divisor can determine that remainder without expanding a long division.

Need help choosing a next step? Scan the contact QR code below or contact Hanlin on WeChat. Send AMC12 REMAINDER ROOTS 1028 and tell us your school year and question.

Write the form of the remainder first

For the original polynomial P(x) = x⁷ + 2x⁴ + 3, find the remainder on division by x² - 1. Write P(x) = (x² - 1)Q(x) + ax + b.

The remainder has degree less than two, so ax + b is the general form. Beginning with only a constant would discard a possibility before using the data.

Use the two zeros of the divisor

At x = 1, the multiple of the divisor vanishes. Since P(1) = 6, we have a + b = 6. At x = -1, P(-1) = 4, giving -a + b = 4.

Adding the equations gives b = 5, and then a = 1. The remainder is x + 5. A single evaluation would leave two unknown coefficients with only one condition.

Check by reducing powers

Modulo x² - 1, we have x² equivalent to 1. Therefore x⁷ reduces to x and x⁴ reduces to 1. The expression reduces to x + 2 + 3 = x + 5.

This is an algebraic congruence modulo a polynomial. It does not assert that x² = 1 for every real value of x.

Notice when the root method needs more care

Two distinct roots provide two independent conditions for a linear remainder. A repeated-root divisor such as (x - 1)² does not give two different conditions merely by substituting x = 1 twice; additional information is needed.

This original AMC 12 exercise is not an official contest problem. In a written solution, state the allowed remainder degree and why each substitution removes the unknown quotient before solving the coefficient equations.

Continue reading: AMC 12 Algebra: Find One Binomial Coefficient Without Expanding Everything.

Contact Hanlin

Planning a first attempt or reviewing a topic you find difficult? Scan the QR code and send AMC12 REMAINDER ROOTS 1028. Include your school year, country or competition region, experience and the question you want to discuss.

Contact Hanlin on WeChat

On your phone, use the button or save the QR image and scan it within WeChat.

Official information and further reading: Source 1