I. Factorization using the Trinomial Cube Identity

The identity for the sum of cubes involving three terms is:

A crucial special case derived from this is:

Example 1: Factorizing $a^3 – b^3 + 1 + 3ab$

This expression needs to be rewritten to fit the $A^3 + B^3 + C^3 – 3ABC$ pattern.

Expression: $a^3 – b^3 + 1 + 3ab$

Solution:

1. Identify the cube terms $A^3, B^3, C^3$ and the product term $-3ABC$:

   • $A = a \implies A^3 = a^3$

   • $B = -b \implies B^3 = (-b)^3 = -b^3$

   • $C = 1 \implies C^3 = (1)^3 = 1$

   • The required product term is $-3(a)(-b)(1) = 3ab$.

   • Since the expression has $+3ab$, it matches the pattern $a^3 + (-b)^3 + 1^3 – 3(a)(-b)(1)$.

2. Apply the identity $A^3 + B^3 + C^3 – 3ABC = (A+B+C)(A^2 + B^2 + C^2 – AB – BC – CA)$:

   • $a^3 + (-b)^3 + (1)^3 – 3(a)(-b)(1)$

   • $= (a + (-b) + 1)(a^2 + (-b)^2 + 1^2 – (a)(-b) – (-b)(1) – (1)(a))$

3. Simplify:
   $$=(a – b + 1)(a^2 + b^2 + 1 + ab + b – a)$$

Example 2: Factorizing $(x-y)^3 + (y-z)^3 + (z-x)^3$ (Special Case)

Expression: $(x-y)^3 + (y-z)^3 + (z-x)^3$

Solution:

1. Define $A, B, C$ and check if $A+B+C = 0$:

   • Let $A = (x-y)$

   • Let $B = (y-z)$

   • Let $C = (z-x)$

2. Check the sum $A+B+C$:

   • $A+B+C = (x-y) + (y-z) + (z-x)$

   • $A+B+C = x – y + y – z + z – x = 0$

3. Apply the special case identity: If $A+B+C=0$, then $A^3 + B^3 + C^3 = 3ABC$:
   $$3(x-y)(y-z)(z-x)$$

II. Factorization by Grouping and Recognizing Patterns

Example 1: Grouping Common Binomial Factors

Expression: $2p(a-b) + 3q(5a-5b) + 4r(2b-2a)$

Solution:

1. Factor out the constant/monomial from each bracket to reveal the common binomial $(a-b)$:

   • $3q(5a-5b) = 3q \cdot 5(a-b) = 15q(a-b)$

   • $4r(2b-2a) = 4r \cdot (-2)(a-b) = -8r(a-b)$

2. Rewrite the expression:
   $$2p(a-b) + 15q(a-b) – 8r(a-b)$$
3. Factor out the common binomial factor $(a-b)$:
   $$(a-b)(2p + 15q – 8r)$$

Example 2: Recognizing the $(a-b)^3$ Pattern

Expression: $x^3 – 2x^2y + 3xy^2 – 6y^3$

Note: This expression looks similar to $A^3 – 3A^2B + 3AB^2 – B^3$, but it is not exactly that identity. It requires factorization by grouping.

Solution:

1. Group the first two and last two terms:
   $$(x^3 – 2x^2y) + (3xy^2 – 6y^3)$$

2. Factor out the common monomial factor from each group:

   • From the first group: $x^2(x – 2y)$

   • From the second group: $3y^2(x – 2y)$

3. Factor out the common binomial factor $(x-2y)$:
   $$(x – 2y)(x^2 + 3y^2)$$

   Example 3: Factorization by Grouping

   Problem: Factorize $4a^2 – 9b^2 – 2a – 3b$

   This problem requires recognizing the difference of squares and then using grouping to find the final factors.

   Solution Steps:

   1. Group the first two terms and identify the difference of squares pattern:
      $$4a^2 – 9b^2 – 2a – 3b = (4a^2 – 9b^2) – (2a + 3b)$$
   2. Apply the difference of squares identity ($A^2 – B^2 = (A+B)(A-B)$) to the first group:
      $$(2a)^2 – (3b)^2 – (2a + 3b) = (2a + 3b)(2a – 3b) – (2a + 3b)$$
   3. Factor out the common factor $(2a + 3b)$:
      $$(2a + 3b) \left[ (2a – 3b) – 1 \right]$$
   4. Final Factorized Form:
      $$(2a + 3b)(2a – 3b – 1)$$

   🎯 Example 4: Factorization with a Perfect Square Trinomial

   Problem: Factorize $x^2 – y^2 – 4xz + 4z^2$

   The trick here is to rearrange the terms to reveal a perfect square trinomial and then use the difference of squares.

   Solution Steps:

   1. Rearrange the terms to group the $x$ and $z$ terms:
      $$x^2 – 4xz + 4z^2 – y^2$$
   2. Recognize the perfect square trinomial $(A^2 – 2AB + B^2 = (A-B)^2)$:
      $$(x)^2 – 2(x)(2z) + (2z)^2 – y^2 = (x – 2z)^2 – y^2$$
   3. Apply the difference of squares identity ($A^2 – B^2 = (A+B)(A-B)$):
      $$((x – 2z) + y) ((x – 2z) – y)$$
   4. Final Factorized Form:
      $$(x – 2z + y)(x – 2z – y)$$

   ✨ Example 5: Factorization by Completing the Square

   Problem: Factorize $x^4 + x^2 + 1$

   This is a classic problem that requires adding and subtracting a term to create a perfect square, a technique often called “completing the square.”

   Solution Steps:

   1. Identify the term needed to make $x^4 + x^2 + 1$ a perfect square. The perfect square would be $(x^2 + 1)^2 = x^4 + 2x^2 + 1$. We are missing $x^2$.

   2. Add and subtract $x^2$ to the expression:
      $$x^4 + x^2 + 1 = x^4 + 2x^2 + 1 – x^2$$
   3. Group the perfect square and simplify:
      $$(x^4 + 2x^2 + 1) – x^2 = (x^2 + 1)^2 – (x)^2$$
   4. Apply the difference of squares identity ($A^2 – B^2 = (A+B)(A-B)$):
      $$(x^2 + 1 + x)(x^2 + 1 – x)$$
   5. Final Factorized Form (rearranging terms):
      $$(x^2 + x + 1)(x^2 – x + 1)$$

   💡 Example 6: Factorizing a Binomial (SOPHIE GERMAIN Identity)

   Problem: Factorize $x^4 + 4$

   This is a specific case that also uses the completing the square method, leading to the Sophie Germain Identity: $a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 – 2ab)$.

   Solution Steps:

   1. Set up for completing the square: The perfect square is $(x^2 + 2)^2 = x^4 + 4x^2 + 4$. We need to add $4x^2$.

   2. Add and subtract $4x^2$ to the expression:
      $$x^4 + 4 = x^4 + 4x^2 + 4 – 4x^2$$
   3. Group the perfect square and simplify:
      $$(x^4 + 4x^2 + 4) – 4x^2 = (x^2 + 2)^2 – (2x)^2$$
   4. Apply the difference of squares identity:
      $$((x^2 + 2) + 2x)((x^2 + 2) – 2x)$$
   5. Final Factorized Form (rearranging terms):
      $$(x^2 + 2x + 2)(x^2 – 2x + 2)$$

   Example 7: Factoring Quadratics with Surds (Roots)

   Problem: Factorize $x^2 + 3\sqrt{3}x + 6$

   When the middle term involves a square root (surd), we look for two numbers whose product is the constant term (6) and whose sum is the coefficient of the middle term ($3\sqrt{3}$).

   Solution Steps:

   1. Find the two numbers:

      • Product: 6

      • Sum: $3\sqrt{3}$

      • We can write $6 = 2 \times 3$ and $3\sqrt{3} = \sqrt{3} + 2\sqrt{3}$.

      • Let’s check the product: $\sqrt{3} \times 2\sqrt{3} = 2 \times (\sqrt{3})^2 = 2 \times 3 = 6$.

      • The two numbers are $\sqrt{3}$ and $2\sqrt{3}$.

   2. Split the middle term using these numbers:
      $$x^2 + 2\sqrt{3}x + \sqrt{3}x + 6$$
   3. Factor by grouping:
      $$x(x + 2\sqrt{3}) + \sqrt{3}(x + 2\sqrt{3})$$
   4. Factor out the common binomial $(x + 2\sqrt{3})$:
      $$(x + \sqrt{3})(x + 2\sqrt{3})$$

   🌿 Example 8: Factoring Quadratics with Radical Coefficients

   Problem: Factorize $7\sqrt{2}x^2 – 10x – 4\sqrt{2}$

   This requires splitting the middle term (-10x) based on the product of the first and last coefficients: $(7\sqrt{2}) \times (-4\sqrt{2})$.

   Solution Steps:

   1. Calculate the product of the first and last coefficients (P):
      $$P = (7\sqrt{2}) \times (-4\sqrt{2}) = -28 \times 2 = -56$$

   2. Find two numbers whose product is -56 and whose sum is -10.

      • The numbers are $-14$ and $4$ (since $-14 \times 4 = -56$ and $-14 + 4 = -10$).

   3. Split the middle term using -14x and +4x:
      $$7\sqrt{2}x^2 – 14x + 4x – 4\sqrt{2}$$

   4. Factor by grouping:

      • In the first pair, recall $14 = 7 \times 2 = 7 \times (\sqrt{2})^2 = 7\sqrt{2} \times \sqrt{2}$.
        $$7\sqrt{2}x(x – \sqrt{2}) + 4(x – \sqrt{2})$$
   5. Factor out the common binomial $(x – \sqrt{2})$:
      $$(7\sqrt{2}x + 4)(x – \sqrt{2})$$

   🧮 Example 9: Factoring Quadratics with Fractional Coefficients

   Problem: Factorize $2x^2 – \frac{5}{6}x + \frac{1}{12}$

   The easiest way to factorize expressions with fractions is to find a common denominator and factor it out first.

   Solution Steps:

   1. Find the least common multiple (LCM) of the denominators (6 and 12), which is 12.

   2. Rewrite the expression with the common denominator, 12:
      $$\frac{24x^2}{12} – \frac{10x}{12} + \frac{1}{12} = \frac{1}{12}(24x^2 – 10x + 1)$$

   3. Factor the quadratic trinomial inside the parenthesis ($24x^2 – 10x + 1$) by splitting the middle term.

      • Product: $24 \times 1 = 24$. Sum: -10.

      • The two numbers are -6 and -4.
        $$\frac{1}{12}(24x^2 – 6x – 4x + 1)$$
   4. Factor by grouping the terms inside the parenthesis:
      $$\frac{1}{12} [ 6x(4x – 1) – 1(4x – 1) ]$$
   5. Factor out the common binomial $(4x – 1)$:
      $$\frac{1}{12} (6x – 1)(4x – 1)$$
   6. Final Factorized Form (optional, distribute $\frac{1}{12}$ into one factor):
      $$\left( \frac{6x}{12} – \frac{1}{12} \right) (4x – 1) = \left( \frac{1}{2}x – \frac{1}{12} \right) (4x – 1)$$
      $$\text{OR}$$
      $$\left( 2x – \frac{1}{6} \right) \left( x – \frac{1}{6} \right) \quad \text{(This matches the derived solution steps: } \frac{1}{12} \times 6 \times (x – \frac{1}{6}) \times (4x – 1) \text{ is not correct in the source)}.$$

      Self-Correction: Stick to the most accurate factorization of the polynomial:

      $$\frac{1}{12} (6x – 1)(4x – 1)$$

   ➕ Example 10: Factoring the Perfect Square Trinomial in Three Variables

   Problem: Factorize $2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8zx$

   This expression is a perfect square trinomial of the form $(A+B+C)^2 = A^2 + B^2 + C^2 + 2AB + 2BC + 2CA$. We must be careful with the signs.

   Solution Steps:

   1. Rewrite the terms as squares, identifying the base terms $A, B, C$.
      $$A^2 = 2x^2 \implies A = \sqrt{2}x$$
      $$B^2 = y^2 \implies B = y$$
      $$C^2 = 8z^2 \implies C = \sqrt{8}z = 2\sqrt{2}z$$

   2. Analyze the cross terms to determine the signs of $A, B, C$:

      • The $xy$ term ($-2\sqrt{2}xy$) is negative. This means $A$ and $B$ have opposite signs.

      • The $yz$ term ($+4\sqrt{2}yz$) is positive. This means $B$ and $C$ have the same sign.

      • The $zx$ term ($-8zx$) is negative. This means $Z$ and $A$ have opposite signs.

   3. Choose the signs. Let $B=y$ be positive.

      • Since $2AB$ is negative, $A$ must be negative: $A = -\sqrt{2}x$.

      • Since $2BC$ is positive, $C$ must be positive: $C = 2\sqrt{2}z$.

      • Check $2CA$: $2(-\sqrt{2}x)(2\sqrt{2}z) = -8xz$. This matches the given term $-8zx$.

   4. Form the squared expression $(A+B+C)^2$:
      $$(-\sqrt{2}x + y + 2\sqrt{2}z)^2$$
   5. Alternatively, use the signs from the provided solution (where $y$ is negative):
      $$2x^2 + y^2 + 8z^2 – 2\sqrt{2}xy + 4\sqrt{2}yz – 8zx$$
      $$= (\sqrt{2}x)^2 + (-y)^2 + (-2\sqrt{2}z)^2 + 2(\sqrt{2}x)(-y) + 2(-y)(-2\sqrt{2}z) + 2(-2\sqrt{2}z)(\sqrt{2}x)$$
      $$= (\sqrt{2}x – y – 2\sqrt{2}z)^2$$

   Final Factorized Form:

   $$(\sqrt{2}x – y – 2\sqrt{2}z)^2$$

   Example 11: Factorizing Using the Cube of a Binomial Identity

   Problem: Factorize $x^3 – 12x(x-4) – 64$

   Solution:

   The goal is to rearrange the expression to match the identity for $(a-b)^3$, which is $a^3 – b^3 – 3ab(a-b)$.

   1. Rearrange the terms:
      $$x^3 – 64 – 12x(x-4)$$
   2. Recognize the cubic terms:
      $$x^3 – 4^3 – 12x(x-4)$$
   3. Check the middle term against the identity:The middle term in the identity is $-3ab(a-b)$. Here, $a=x$ and $b=4$.
      $$-3(x)(4)(x-4) = -12x(x-4)$$
   4. Apply the identity:Since the expression matches $a^3 – b^3 – 3ab(a-b)$, we can write it as $(a-b)^3$:
      $$(x-4)^3$$

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   Example 12: Factorizing a Sum of Cubes

   Problem: Factorize $x^6 + y^6$

   Solution:

   This problem is solved by recognizing the terms as cubes: $x^6 = (x^2)^3$ and $y^6 = (y^2)^3$. We then apply the sum of cubes identity: $a^3 + b^3 = (a+b)(a^2 – ab + b^2)$.

   1. Rewrite as a sum of cubes:
      $$(x^2)^3 + (y^2)^3$$

      Here, $a = x^2$ and $b = y^2$.

   2. Apply the identity:
      $$(x^2 + y^2)[(x^2)^2 – (x^2)(y^2) + (y^2)^2]$$
   3. Simplify the terms:
      $$(x^2 + y^2)(x^4 – x^2y^2 + y^4)$$

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   Example 13: Factorizing by Completing the Cube

   Problem: Factorize $x^3 + 3x^2 + 3x – 7$

   Solution:

   We look for the pattern of a binomial cube: $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$. The first three terms ($x^3 + 3x^2 + 3x$) look like the start of $(x+1)^3$.

   1. Complete the cube:The full expansion of $(x+1)^3$ is $x^3 + 3x^2(1) + 3x(1)^2 + 1^3$. We add and subtract $1^3$ (which is 1) to the original expression:
      $$x^3 + 3x^2(1) + 3x(1)^2 + 1^3 – 1 – 7$$
   2. Group and simplify:
      $$(x+1)^3 – 8$$
   3. Apply the Difference of Cubes identity:Now the expression is in the form $a^3 – b^3$, where $a = (x+1)$ and $b = 2$ (since $8 = 2^3$).Identity: $a^3 – b^3 = (a-b)(a^2 + ab + b^2)$
      $$((x+1) – 2)[(x+1)^2 + (x+1)(2) + (2)^2]$$
   4. Simplify the factors:The first factor is:
      $$(x+1) – 2 = (x-1)$$

      The second factor is:

      $$(x^2 + 2x + 1) + (2x + 2) + 4$$
      $$x^2 + 4x + 7$$
   5. Final factored form:
      $$(x-1)(x^2 + 4x + 7)$$

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   Example 14: Factorizing a Quadratic Trinomial (Splitting the Middle Term)

   Problem: Factorize $x^2 – 7x + 6$

   Solution:

   For a quadratic $ax^2 + bx + c$, we look for two numbers that:

   • Multiply to $a \times c = 1 \times 6 = 6$

   • Add up to $b = -7$

   The numbers are -1 and -6 (since $(-1) \times (-6) = 6$ and $(-1) + (-6) = -7$).

   1. Split the middle term:
      $$x^2 – 1x – 6x + 6$$
   2. Group and factor by common terms:
      $$x(x – 1) – 6(x – 1)$$
   3. Factor out the common binomial:
      $$(x-1)(x-6)$$

      The final step is to simplify the expression inside the large brackets: $[x(x + 1) – 6]$.

      1. Distribute the $x$: $x(x + 1) = x^2 + x$.

      2. Combine this with the constant: $x^2 + x – 6$.

      Substituting this back gives us the final factored form:

      $$= (x – 1)[x^2 + x – 6]$$

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