9.2 Algebraic Identities (Çarpanlara Ayırma), Sums and Differences of Cubes & Reciprocal Sums

Algebraic Identities & Factoring (Çarpanlara Ayırma)

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Fundamental Factoring Methods & Difference of Squares

1. Common Factor Extraction & 4-Term Grouping (Gruplandırma Yöntemi)

• Sign-Reversal Factoring: In rational fractions, always watch for opposite binomials: $(b - a) = -(a - b)$, whereas under an even exponent $(b - a)^2 = (a - b)^2$.
• Grouping by Pairs (Gruplandırma): When a four-term polynomial has no single factor shared by all four terms, group the terms into two pairs that share a common binomial:

$$ax + ay + bx + by = a(x + y) + b(x + y) = (a + b)(x + y)$$

• Quadratic Trinomial Factoring: To factor $x^2 + bx + c$, find two numbers $m, n$ with product $m \cdot n = c$ and sum $m + n = b$, giving $(x + m)(x + n)$. For $ax^2 + bx + c$ ($a \ne 1$), decompose $a = p \cdot q$ and $c = r \cdot s$ so that the cross-product sum satisfies $p s + q r = b$, yielding $(px + r)(qx + s)$.

2. Difference of Squares (İki Kare Farkı): $a^2 - b^2 = (a - b)(a + b)$

Beyond basic algebraic factoring, TR-YÖS tests three specialized applications of $a^2 - b^2$:

1. Telescoping Dyadic Products: Multiplying $(x - 1)$ by successive powers $(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1)$ collapses step-by-step via difference of squares:

$$(x - 1)(x + 1)(x^2 + 1)(x^4 + 1)(x^8 + 1) = (x^2 - 1)(x^2 + 1)(x^4 + 1)(x^8 + 1) = x^{16} - 1$$

1. Large-Number Radical Arithmetic: To evaluate $\sqrt{2026 \times 2022 + 4}$ without multiplying four-digit numbers, let the midpoint be $m = 2024$:

$$\sqrt{(2024 + 2)(2024 - 2) + 4} = \sqrt{(2024^2 - 2^2) + 4} = \sqrt{2024^2} = 2024$$

1. Prime-Difference Integer Equations: If $a, b \in \mathbb{Z}^+$ and $a^2 - b^2 = p$ where $p$ is an odd positive prime number, then $(a - b)(a + b) = 1 \cdot p$. Since $a + b > a - b > 0$ for positive integers, the factors are uniquely forced:

$$a - b = 1 \quad \text{and} \quad a + b = p \implies a = \frac{p + 1}{2}, \quad b = \frac{p - 1}{2}$$

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Perfect Square & Cubic Identities (Tam Kare ve Küp Özdeşlikleri)

1. Perfect Square Identities (Tam Kare Özdeşlikleri)

$$(a \pm b)^2 = a^2 \pm 2ab + b^2$$

Rearranging this expansion yields the value-conversion identities that connect $a^2 + b^2$, $a + b$, $a - b$, and $ab$:

$$a^2 + b^2 = (a + b)^2 - 2ab = (a - b)^2 + 2ab, \qquad (a + b)^2 - (a - b)^2 = 4ab$$

For three variables, the square of a trinomial expands to the sum of the squares plus twice every pairwise product:

$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ac)$$

If any term carries a minus sign—such as $(a + b - c)^2$—any pairwise product involving $-c$ becomes negative: $(a + b - c)^2 = a^2 + b^2 + c^2 + 2(ab - ac - bc)$.

2. Sum & Difference of Cubes (İki Küp Toplamı ve Farkı) vs. Cube of a Binomial

First, distinguish the cube of a binomial $(a \pm b)^3$ from the sum or difference of two cubes $a^3 \pm b^3$:

$$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = a^3 + b^3 + 3ab(a + b)$$ $$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a - b)$$

For $a^3 \pm b^3$, you must know both algebraic forms and when to deploy each:

• Form 1 — Factoring Form (use when simplifying algebraic fractions):

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2), \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$

Exam Tip: Notice that the quadratic factor $a^2 \mp ab + b^2$ has coefficient $\mp 1$ on $ab$ (not $\mp 2$!). It never factors further over $\mathbb{R}$, and in rational simplification problems it almost always cancels an identical unfactorable quadratic in the denominator.

• Form 2 — Value-Computation Form (use when numerical values of $a \pm b$ and $ab$ are given):

$$a^3 + b^3 = (a + b)^3 - 3ab(a + b), \qquad a^3 - b^3 = (a - b)^3 + 3ab(a - b)$$

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The Reciprocal Power Ladder ($x \pm \frac{1}{x}$ Kalıpları)

$$x^2 - kx + 1 = 0 \xrightarrow{\div x} x - k + \frac{1}{x} = 0 \implies x + \frac{1}{x} = k$$

Since the product of $x$ and $\frac{1}{x}$ is $x \cdot \frac{1}{x} = 1$, substituting $a = x$ and $b = \frac{1}{x}$ into the square and cube value-computation formulas makes every $ab$ factor disappear:

$$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = \left(x - \frac{1}{x}\right)^2 + 2$$ $$\left(x + \frac{1}{x}\right)^2 - \left(x - \frac{1}{x}\right)^2 = 4$$ $$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right), \qquad x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right)$$

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Sum-of-Squares Equals Zero & Completing the Square for Extrema

1. Sum of Squares Equals Zero (Kareler Toplamının Sıfıra Eşitliği)

Because the square of any real number is non-negative ($u^2 \ge 0$), a sum of squares of real expressions can equal zero if and only if every base equals zero simultaneously:

$$(x - a)^2 + (y - b)^2 = 0 \iff x = a \quad \text{and} \quad y = b \qquad (x, y \in \mathbb{R})$$

On TR-YÖS, this equation is given in expanded form, such as $x^2 + y^2 - 6x + 8y + 25 = 0$. Split the constant $25 = 9 + 16$ to complete both squares: $(x^2 - 6x + 9) + (y^2 + 8y + 16) = (x - 3)^2 + (y + 4)^2 = 0$, which forces $x = 3$ and $y = -4$.

2. Completing the Square for Minimum / Maximum Values

To find the minimum value of a quadratic $P(x) = x^2 + 2bx + c$ over $x \in \mathbb{R}$, add and subtract $b^2$ (half the linear coefficient squared):

$$P(x) = (x^2 + 2bx + b^2) + c - b^2 = (x + b)^2 + (c - b^2)$$

Since $(x + b)^2 \ge 0$ for all $x \in \mathbb{R}$, the minimum value of $P(x)$ is $c - b^2$, attained when $x = -b$.

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Reference Table of Core Algebraic Identities

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Worked Examples

Worked Example 1: Climbing the Reciprocal Power Ladder

Problem: Given that $x^2 - 4x + 1 = 0$, find the values of (a) $x^3 + \frac{1}{x^3}$ and (b) $x^4 + \frac{1}{x^4}$.

Solution:

1. Since $x \ne 0$, divide every term of $x^2 - 4x + 1 = 0$ by $x$:

$$x - 4 + \frac{1}{x} = 0 \implies x + \frac{1}{x} = 4$$

1. (a) Apply the reciprocal cube identity with $k = 4$:

$$x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) = 4^3 - 3(4) = 64 - 12 = \mathbf{52}$$

1. (b) First compute $x^2 + \frac{1}{x^2}$, then square a second time to reach the fourth powers:

$$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = 4^2 - 2 = 14$$ $$x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2 = 14^2 - 2 = 196 - 2 = \mathbf{194}$$

Worked Example 2: Rational Simplification via Difference of Cubes

Problem: Simplify $\frac{x^3 - 8}{x^2 - 4} \div \frac{x^2 + 2x + 4}{x^2 + 3x + 2}$ for $x \notin \{-2, -1, 2\}$.

Solution:

1. Factor each polynomial using difference of cubes $x^3 - 2^3$, difference of squares $x^2 - 2^2$, and trinomial factoring:
   • $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$
   • $x^2 - 4 = (x - 2)(x + 2)$
   • $x^2 + 3x + 2 = (x + 2)(x + 1)$
2. Invert the divisor into a product and cancel matching factors:

$$\frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 2)} \cdot \frac{(x + 2)(x + 1)}{x^2 + 2x + 4} = \mathbf{x + 1}$$

Review the Method

Completing the square rewrites $x^2 + 2bx + c$ as $(x + b)^2 + (c - b^2)$, immediately exposing its minimum value $c - b^2$ at $x = -b$.

Cancellation Does Not Restore an Excluded Input

Factoring changes the form of an expression, but it does not erase the original denominator restriction. Before cancelling, write the forbidden inputs beside your work. After cancellation, check every proposed value against that list. This matters especially when the simplified result looks like a polynomial that can be evaluated everywhere.

For example, consider $F(x)=\frac{x^2-1}{x-1}$. The original expression requires $x\ne1$. Factoring the numerator gives $(x-1)(x+1)$, so cancellation yields $F(x)=x+1$ only for the original allowed inputs. At $x=2$, either form gives 3. At $x=1$, the simplified polynomial gives 2, but the original fraction is undefined. Its graph therefore has a missing point at that input rather than becoming the complete line.

Now solve $\frac{x^2-1}{x-1}=2$. Record $x\ne1$, simplify to $x+1=2$, and obtain the candidate $x=1$. Reject that candidate because it violates the original restriction: this equation has no solution. Substitution into the original expression is the final check, even when the intermediate algebra is correct.

The same discipline applies when dividing by an algebraic fraction. Its denominator must be nonzero, and the fraction itself must also be nonzero because it is the divisor. Inverting it without recording both restrictions can silently introduce invalid values. In the worked difference-of-cubes example above, the stated excluded values come from the original denominators; keep them after the factors cancel.

Finally, distinguish an identity from an equation with selected solutions. The difference-of-squares identity holds for every pair of real inputs. An equation built using it may have only a few permitted roots after domain restrictions are imposed. Explain both the algebraic simplification and the allowed input set when checking an answer.