OP Malhotra Class 9 Maths Solutions Chapter 4 Factorisation Exercise 4 (H)

S Chand Class 9 ICSE Maths Solutions Chapter 4 Factorisation Ex 4(H)

Factorise the following:

Question 1. \( a^3 + 1 \)
Answer: We can write \( a^3 + 1 \) as \( (a)^3 + (1)^3 \). This is a sum of cubes, which follows the formula \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \). So, when we apply this formula, we get \( (a + 1)(a^2 - a \cdot 1 + 1^2) \), which simplifies to \( (a + 1)(a^2 - a + 1) \). This helps break down complex expressions into simpler parts.
\( a^3 + 1 = (a)^3 + (1)^3 \)
\( \implies (a + 1)(a^2 - a \times 1 + 1^2) \)
\( \implies (a + 1)(a^2 - a + 1) \)
In simple words: To factor \( a^3 + 1 \), think of it as two cubes being added. Use the special formula for adding cubes to split it into two brackets.

🎯 Exam Tip: Remember the sum of cubes formula: \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \). It's crucial for factoring these types of expressions.

Question 2. \( x^3 + 8 \)
Answer: We can express \( x^3 + 8 \) as \( (x)^3 + (2)^3 \). This is another sum of cubes, where \( x \) is \( a \) and \( 2 \) is \( b \) in the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). By using this formula, the expression becomes \( (x + 2)((x)^2 - x \cdot 2 + (2)^2) \), which simplifies to \( (x + 2)(x^2 - 2x + 4) \). This method simplifies the given polynomial into a product of simpler factors.
\( x^3 + 8 = (x)^3 + (2)^3 \)
\( \implies (x + 2)[(x)^2 - x \times 2 + (2)^2] \)
\( \implies (x + 2)(x^2 - 2x + 4) \)
In simple words: Think of \( x^3 + 8 \) as \( x \) cubed plus \( 2 \) cubed. Use the sum of cubes rule to break it into two parts in brackets.

🎯 Exam Tip: Always identify if the expression fits a standard factorization identity like sum or difference of cubes. This makes factoring much easier.

Question 3. \( 8x^3 + 1 \)
Answer: We can rewrite \( 8x^3 + 1 \) as \( (2x)^3 + (1)^3 \). This is a sum of cubes, using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), where \( a \) is \( 2x \) and \( b \) is \( 1 \). When we apply the formula, it expands to \( (2x + 1)((2x)^2 - 2x \cdot 1 + (1)^2) \), which simplifies to \( (2x + 1)(4x^2 - 2x + 1) \). This technique turns a cube sum into a product of a linear and a quadratic factor.
\( 8x^3 + 1 = (2x)^3 + (1)^3 \)
\( \implies (2x + 1)[(2x)^2 - 2x \times 1 + (1)^2] \)
\( \implies (2x + 1)(4x^2 - 2x + 1) \)
In simple words: For \( 8x^3 + 1 \), see it as \( 2x \) cubed plus \( 1 \) cubed. Then apply the sum of cubes formula to factor it.

🎯 Exam Tip: For terms like \( 8x^3 \), always find the cube root of the coefficient (like 8) and the variable (like \( x^3 \)) to correctly identify the 'a' term, which here is \( 2x \).

Question 4. \( x^3 - 27 \)
Answer: We can write \( x^3 - 27 \) as \( (x)^3 - (3)^3 \). This is a difference of cubes, which follows the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Applying this identity, the expression becomes \( (x - 3)((x)^2 + x \cdot 3 + (3)^2) \), which simplifies to \( (x - 3)(x^2 + 3x + 9) \). This factorization is a fundamental algebraic manipulation.
\( x^3 - 27 = (x)^3 - (3)^3 \)
\( \implies (x - 3)[(x)^2 + x \times 3 + (3)^2] \)
\( \implies (x - 3)(x^2 + 3x + 9) \)
In simple words: To factor \( x^3 - 27 \), think of it as \( x \) cubed minus \( 3 \) cubed. Use the special rule for subtracting cubes to split it.

🎯 Exam Tip: The difference of cubes formula is \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). Notice the signs are different from the sum of cubes formula.

Question 5. \( a^3 - 8 \)
Answer: We can write \( a^3 - 8 \) as \( (a)^3 - (2)^3 \). This is a difference of cubes, using the formula \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \), where \( x \) is \( a \) and \( y \) is \( 2 \). Applying this formula, the expression becomes \( (a - 2)((a)^2 + a \cdot 2 + (2)^2) \), which simplifies to \( (a - 2)(a^2 + 2a + 4) \). Factoring helps in solving equations and simplifying rational expressions.
\( a^3 - 8 = (a)^3 - (2)^3 \)
\( \implies (a - 2)[(a)^2 + a \times 2 + (2)^2] \)
\( \implies (a - 2)(a^2 + 2a + 4) \)
In simple words: See \( a^3 - 8 \) as \( a \) cubed minus \( 2 \) cubed. Then use the difference of cubes rule to break it into two sets of brackets.

🎯 Exam Tip: Always double-check your signs in the quadratic factor \( (a^2 + ab + b^2) \). For a difference of cubes, all terms in this factor are positive.

Question 6. \( 27m^3 - 8 \)
Answer: We can rewrite \( 27m^3 - 8 \) as \( (3m)^3 - (2)^3 \). This fits the difference of cubes identity \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \), where \( A \) is \( 3m \) and \( B \) is \( 2 \). Applying this identity, the expression becomes \( (3m - 2)((3m)^2 + 3m \cdot 2 + (2)^2) \), which simplifies to \( (3m - 2)(9m^2 + 6m + 4) \). This factorization is an important skill in algebra.
\( 27m^3 - 8 = (3m)^3 - (2)^3 \)
\( \implies (3m - 2)[(3m)^2 + 3m \times 2 + (2)^2] \)
\( \implies (3m - 2)(9m^2 + 6m + 4) \)
In simple words: For \( 27m^3 - 8 \), think of it as \( 3m \) cubed minus \( 2 \) cubed. Then apply the difference of cubes formula to factor it.

🎯 Exam Tip: Ensure that when you square a term like \( 3m \), both the coefficient and the variable are squared, resulting in \( 9m^2 \).

Question 7. \( x^3 + 64 \)
Answer: We can express \( x^3 + 64 \) as \( (x)^3 + (4)^3 \). This is a sum of cubes, using the formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Here, \( a \) is \( x \) and \( b \) is \( 4 \). Applying the formula, it expands to \( (x + 4)((x)^2 - x \cdot 4 + (4)^2) \), which simplifies to \( (x + 4)(x^2 - 4x + 16) \). Recognizing perfect cubes is the first step in these problems.
\( x^3 + 64 = (x)^3 + (4)^3 \)
\( \implies (x + 4)[(x)^2 - x \times 4 + (4)^2] \)
\( \implies (x + 4)(x^2 - 4x + 16) \)
In simple words: Think of \( x^3 + 64 \) as \( x \) cubed plus \( 4 \) cubed. Then use the sum of cubes rule to break it into two factors.

🎯 Exam Tip: Knowing common perfect cubes (like \( 64 = 4^3 \)) will speed up your factorization process. Always check if the constant term is a perfect cube.

Question 8. \( 8a^3 - b^6 \)
Answer: We can write \( 8a^3 - b^6 \) as \( (2a)^3 - (b^2)^3 \). This is a difference of cubes, following the identity \( X^3 - Y^3 = (X - Y)(X^2 + XY + Y^2) \), where \( X \) is \( 2a \) and \( Y \) is \( b^2 \). Applying this identity, the expression becomes \( (2a - b^2)((2a)^2 + 2a \cdot b^2 + (b^2)^2) \), which simplifies to \( (2a - b^2)(4a^2 + 2ab^2 + b^4) \). This shows how the difference of cubes applies even to terms with exponents.
\( 8a^3 - b^6 = (2a)^3 - (b^2)^3 \)
\( \implies (2a - b^2)[(2a)^2 + 2a \times b^2 + (b^2)^2] \)
\( \implies (2a - b^2)(4a^2 + 2ab^2 + b^4) \)
In simple words: See \( 8a^3 - b^6 \) as \( 2a \) cubed minus \( b^2 \) cubed. Use the difference of cubes rule to factor it.

🎯 Exam Tip: When dealing with variables raised to higher powers (like \( b^6 \)), remember that \( (b^2)^3 = b^6 \). This helps identify the correct terms for cube formulas.

Question 9. \( x^6 + 8b^3 \)
Answer: We can express \( x^6 + 8b^3 \) as \( (x^2)^3 + (2b)^3 \). This fits the sum of cubes identity \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \), where \( A \) is \( x^2 \) and \( B \) is \( 2b \). Using this formula, the expression expands to \( (x^2 + 2b)((x^2)^2 - x^2 \cdot 2b + (2b)^2) \), which simplifies to \( (x^2 + 2b)(x^4 - 2x^2b + 4b^2) \). This demonstrates factoring with terms that are already powers.
\( x^6 + 8b^3 = (x^2)^3 + (2b)^3 \)
\( \implies (x^2 + 2b)[(x^2)^2 - x^2 \times 2b + (2b)^2] \)
\( \implies (x^2 + 2b)(x^4 - 2x^2b + 4b^2) \)
In simple words: For \( x^6 + 8b^3 \), imagine it as \( x^2 \) cubed plus \( 2b \) cubed. Then use the sum of cubes formula to factor it.

🎯 Exam Tip: When a power is raised to another power, like \( (x^2)^2 \), multiply the exponents to get \( x^4 \).

Question 10. \( 8a^3 + 27b^3 \)
Answer: We can rewrite \( 8a^3 + 27b^3 \) as \( (2a)^3 + (3b)^3 \). This is a sum of cubes, using the formula \( X^3 + Y^3 = (X + Y)(X^2 - XY + Y^2) \). Here, \( X \) is \( 2a \) and \( Y \) is \( 3b \). Applying the formula, the expression becomes \( (2a + 3b)((2a)^2 - 2a \cdot 3b + (3b)^2) \), which simplifies to \( (2a + 3b)(4a^2 - 6ab + 9b^2) \). This factorization is key for simplifying more complex algebraic problems later on.
\( 8a^3 + 27b^3 = (2a)^3 + (3b)^3 \)
\( \implies (2a + 3b)[(2a)^2 - 2a \times 3b + (3b)^2] \)
\( \implies (2a + 3b)(4a^2 - 6ab + 9b^2) \)
In simple words: Think of \( 8a^3 + 27b^3 \) as \( 2a \) cubed plus \( 3b \) cubed. Then use the sum of cubes rule to break it into two parts.

🎯 Exam Tip: Ensure that you correctly identify the cube roots of both the coefficients and the variables for each term before applying the formula.

Question 11. \( 27x^3 - 8y^3 \)
Answer: We can write \( 27x^3 - 8y^3 \) as \( (3x)^3 - (2y)^3 \). This is a difference of cubes, which follows the identity \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \). Here, \( A \) is \( 3x \) and \( B \) is \( 2y \). Applying this formula, the expression becomes \( (3x - 2y)((3x)^2 + 3x \cdot 2y + (2y)^2) \), which simplifies to \( (3x - 2y)(9x^2 + 6xy + 4y^2) \). This factorization can be useful in calculus and other advanced math topics.
\( 27x^3 - 8y^3 = (3x)^3 - (2y)^3 \)
\( \implies (3x - 2y)[(3x)^2 + 3x \times 2y + (2y)^2] \)
\( \implies (3x - 2y)(9x^2 + 6xy + 4y^2) \)
In simple words: See \( 27x^3 - 8y^3 \) as \( 3x \) cubed minus \( 2y \) cubed. Then use the difference of cubes rule to factor it.

🎯 Exam Tip: Pay close attention to the signs in the factorization. For \( A^3 - B^3 \), the first bracket is \( (A-B) \), and all terms in the second bracket are positive.

Question 12. \( 128x^3 + 2 \)
Answer: First, we should always look for a common factor. We can factor out \( 2 \) from \( 128x^3 + 2 \), which gives us \( 2(64x^3 + 1) \). Now, the term inside the bracket, \( 64x^3 + 1 \), can be written as \( (4x)^3 + (1)^3 \). This is a sum of cubes, \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \), where \( A \) is \( 4x \) and \( B \) is \( 1 \). Applying the formula, the expression becomes \( 2(4x + 1)((4x)^2 - 4x \cdot 1 + (1)^2) \), which simplifies to \( 2(4x + 1)(16x^2 - 4x + 1) \). Always factor out common terms first to simplify the problem.
\( 128x^3 + 2 = 2 (64x^3 + 1) \)
\( \implies 2 [(4x)^3 + (1)^3] \)
\( \implies 2 (4x + 1)[(4x)^2 - 4x \times 1 + (1)^2] \)
\( \implies 2 (4x + 1)(16x^2 - 4x + 1) \)
In simple words: First, take out the common number 2. Then, the part inside the bracket is \( 4x \) cubed plus \( 1 \) cubed. Use the sum of cubes rule with this part.

🎯 Exam Tip: Always remember to look for a common factor (like 2 in this case) before applying any other factorization identities. This often simplifies the problem significantly.

Question 13. Factorise: \( 8x^3 - \frac{1}{27}y^3 \)
Answer: We can rewrite \( 8x^3 - \frac{1}{27}y^3 \) as \( (2x)^3 - (\frac{1}{3}y)^3 \). This is a difference of cubes, following the identity \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \), where \( A \) is \( 2x \) and \( B \) is \( \frac{1}{3}y \). Applying the formula, the expression becomes \( (2x - \frac{1}{3}y)((2x)^2 + 2x \cdot \frac{1}{3}y + (\frac{1}{3}y)^2) \), which simplifies to \( (2x - \frac{1}{3}y)(4x^2 + \frac{2}{3}xy + \frac{1}{9}y^2) \). This shows how to factor expressions with fractions as coefficients.
\( 8x^3 - \frac{1}{27}y^3 = (2x)^3 - (\frac{1}{3}y)^3 \)
\( \implies (2x - \frac{1}{3}y)[(2x)^2 + 2x \times \frac{1}{3}y + (\frac{1}{3}y)^2] \)
\( \implies (2x - \frac{1}{3}y)(4x^2 + \frac{2}{3}xy + \frac{1}{9}y^2) \)
In simple words: Treat \( 8x^3 \) as \( 2x \) cubed, and \( \frac{1}{27}y^3 \) as \( \frac{1}{3}y \) cubed. Then use the difference of cubes rule to factor it.

🎯 Exam Tip: When dealing with fractions, remember to cube both the numerator and the denominator (e.g., \( (\frac{1}{3})^3 = \frac{1}{27} \)).

Question 14. \( 343x^3y + 512y^4 \)
Answer: First, we need to find a common factor. Both terms have \( y \), so we can factor out \( y \), which gives \( y(343x^3 + 512y^3) \). Now, the expression inside the bracket is a sum of cubes: \( (7x)^3 + (8y)^3 \). We apply the formula \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \), where \( A \) is \( 7x \) and \( B \) is \( 8y \). So, the full factorization becomes \( y(7x + 8y)((7x)^2 - 7x \cdot 8y + (8y)^2) \), which simplifies to \( y(7x + 8y)(49x^2 - 56xy + 64y^2) \). This multi-step factorization is a good way to practice finding common factors and using identities.
\( 343x^3y + 512y^4 = y(343x^3 + 512y^3) \)
\( \implies y[(7x)^3 + (8y)^3] \)
\( \implies y [(7x + 8y) [(7x)^2 - 7x \times 8y + (8y)^2]] \)
\( \implies y (7x + 8y) (49x^2 - 56xy + 64y^2) \)
In simple words: First, take out the common letter \( y \). Then, inside the bracket, recognize that \( 343x^3 \) is \( 7x \) cubed and \( 512y^3 \) is \( 8y \) cubed. Use the sum of cubes rule on these.

🎯 Exam Tip: Always start by looking for a common factor among all terms. This simplifies the expression and makes it easier to spot other factorization patterns.

Question 15. \( (2a + b)^3 + (a + 2b)^3 \)
Answer: This expression is in the form \( X^3 + Y^3 \), where \( X = (2a + b) \) and \( Y = (a + 2b) \). We use the sum of cubes formula: \( X^3 + Y^3 = (X + Y)(X^2 - XY + Y^2) \). First, find \( X + Y \): \( (2a + b) + (a + 2b) = 3a + 3b = 3(a + b) \). Next, find \( X^2 - XY + Y^2 \): \( X^2 = (2a + b)^2 = 4a^2 + 4ab + b^2 \) \( Y^2 = (a + 2b)^2 = a^2 + 4ab + 4b^2 \) \( XY = (2a + b)(a + 2b) = 2a^2 + 4ab + ab + 2b^2 = 2a^2 + 5ab + 2b^2 \) So, \( X^2 - XY + Y^2 = (4a^2 + 4ab + b^2) - (2a^2 + 5ab + 2b^2) + (a^2 + 4ab + 4b^2) \) \( = (4a^2 - 2a^2 + a^2) + (4ab - 5ab + 4ab) + (b^2 - 2b^2 + 4b^2) \) \( = 3a^2 + 3ab + 3b^2 = 3(a^2 + ab + b^2) \). Combining these, the factorization is \( 3(a + b) \cdot 3(a^2 + ab + b^2) = 9(a + b)(a^2 + ab + b^2) \). Breaking down complex terms into simpler variables helps manage the expansion.
Let \( X = 2a + b \) and \( Y = a + 2b \)
Then, \( X^3 + Y^3 = (X + Y)(X^2 - XY + Y^2) \)
\( X + Y = (2a + b) + (a + 2b) = 3a + 3b = 3(a + b) \)
\( X^2 = (2a + b)^2 = 4a^2 + 4ab + b^2 \)
\( Y^2 = (a + 2b)^2 = a^2 + 4ab + 4b^2 \)
\( XY = (2a + b)(a + 2b) = 2a^2 + 5ab + 2b^2 \)
\( X^2 - XY + Y^2 = (4a^2 + 4ab + b^2) - (2a^2 + 5ab + 2b^2) + (a^2 + 4ab + 4b^2) \)
\( \implies (4a^2 - 2a^2 + a^2) + (4ab - 5ab + 4ab) + (b^2 - 2b^2 + 4b^2) \)
\( \implies 3a^2 + 3ab + 3b^2 = 3(a^2 + ab + b^2) \)
Therefore, \( (2a + b)^3 + (a + 2b)^3 = 3(a + b) \cdot 3(a^2 + ab + b^2) \)
\( \implies 9(a + b)(a^2 + ab + b^2) \)
In simple words: Call the first bracket \( X \) and the second bracket \( Y \). Then use the sum of cubes formula on \( X^3 + Y^3 \). Add \( X \) and \( Y \) first, then calculate \( X^2 - XY + Y^2 \), and multiply the results.

🎯 Exam Tip: When terms are expressions themselves (like \( 2a+b \)), substitute them carefully into the identity. Using temporary variables (X, Y) can help avoid errors.

Question 16. \( 27(m + 2n)^3 + (m - 6n)^3 \)
Answer: This expression can be written as \( (3(m + 2n))^3 + (m - 6n)^3 \), which is a sum of cubes, \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \). Let \( A = 3(m + 2n) = 3m + 6n \) and \( B = m - 6n \). First, find \( A + B \): \( (3m + 6n) + (m - 6n) = 4m \). Next, find \( A^2 - AB + B^2 \): \( A^2 = (3m + 6n)^2 = 9m^2 + 36mn + 36n^2 \) \( B^2 = (m - 6n)^2 = m^2 - 12mn + 36n^2 \) \( AB = (3m + 6n)(m - 6n) = 3m^2 - 18mn + 6mn - 36n^2 = 3m^2 - 12mn - 36n^2 \) So, \( A^2 - AB + B^2 = (9m^2 + 36mn + 36n^2) - (3m^2 - 12mn - 36n^2) + (m^2 - 12mn + 36n^2) \) \( = 9m^2 + 36mn + 36n^2 - 3m^2 + 12mn + 36n^2 + m^2 - 12mn + 36n^2 \) \( = (9m^2 - 3m^2 + m^2) + (36mn + 12mn - 12mn) + (36n^2 + 36n^2 + 36n^2) \) \( = 7m^2 + 36mn + 108n^2 \). Combining these, the factorization is \( (4m)(7m^2 + 36mn + 108n^2) \). Breaking down into \( A \) and \( B \) terms simplifies the work.
Let \( A = 3(m + 2n) = 3m + 6n \) and \( B = m - 6n \)
The expression is \( A^3 + B^3 \)
\( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \)
\( A + B = (3m + 6n) + (m - 6n) = 4m \)
\( A^2 = (3m + 6n)^2 = 9m^2 + 36mn + 36n^2 \)
\( B^2 = (m - 6n)^2 = m^2 - 12mn + 36n^2 \)
\( AB = (3m + 6n)(m - 6n) = 3m^2 - 18mn + 6mn - 36n^2 = 3m^2 - 12mn - 36n^2 \)
\( A^2 - AB + B^2 = (9m^2 + 36mn + 36n^2) - (3m^2 - 12mn - 36n^2) + (m^2 - 12mn + 36n^2) \)
\( \implies 9m^2 + 36mn + 36n^2 - 3m^2 + 12mn + 36n^2 + m^2 - 12mn + 36n^2 \)
\( \implies (9-3+1)m^2 + (36+12-12)mn + (36+36+36)n^2 \)
\( \implies 7m^2 + 36mn + 108n^2 \)
Therefore, \( 27(m + 2n)^3 + (m - 6n)^3 = (4m)(7m^2 + 36mn + 108n^2) \)
In simple words: See \( 27(m+2n)^3 \) as \( (3(m+2n))^3 \). Let \( A = 3(m+2n) \) and \( B = (m-6n) \). Use the sum of cubes formula for \( A^3+B^3 \). Add \( A \) and \( B \) first, then calculate \( A^2 - AB + B^2 \), and multiply the two results.

🎯 Exam Tip: When a coefficient is present, like 27 in \( 27(m+2n)^3 \), make sure it is included inside the cubed term as \( (3(m+2n))^3 \). This is a common point of error.

Question 17. \( 8(a + b)^3 - 27c^3 \)
Answer: We can rewrite \( 8(a + b)^3 - 27c^3 \) as \( (2(a + b))^3 - (3c)^3 \). This is a difference of cubes, following the identity \( X^3 - Y^3 = (X - Y)(X^2 + XY + Y^2) \). Let \( X = 2(a + b) = 2a + 2b \) and \( Y = 3c \). First, find \( X - Y \): \( (2a + 2b) - 3c \). Next, find \( X^2 + XY + Y^2 \): \( X^2 = (2a + 2b)^2 = 4(a + b)^2 = 4(a^2 + 2ab + b^2) = 4a^2 + 8ab + 4b^2 \) \( Y^2 = (3c)^2 = 9c^2 \) \( XY = (2a + 2b)(3c) = 6ac + 6bc \) So, \( X^2 + XY + Y^2 = (4a^2 + 8ab + 4b^2) + (6ac + 6bc) + (9c^2) \). Combining these, the factorization is \( (2a + 2b - 3c)(4a^2 + 4b^2 + 9c^2 + 8ab + 6ac + 6bc) \). Remember to expand all terms fully for the final answer.
\( 8(a + b)^3 - 27c^3 = [2(a + b)]^3 - (3c)^3 \)
\( \implies [2(a + b) - 3c][\{2(a + b)\}^2 + 2(a + b) \times 3c + (3c)^2] \)
\( \implies (2a + 2b - 3c)[4(a^2 + 2ab + b^2) + 6ac + 6bc + 9c^2] \)
\( \implies (2a + 2b - 3c)[4a^2 + 8ab + 4b^2 + 6ac + 6bc + 9c^2] \)
\( \implies (2a + 2b - 3c)(4a^2 + 4b^2 + 9c^2 + 8ab + 6ac + 6bc) \)
In simple words: This is a difference of cubes. Think of \( 8(a+b)^3 \) as \( (2(a+b))^3 \) and \( 27c^3 \) as \( (3c)^3 \). Apply the difference of cubes formula by substituting \( 2(a+b) \) and \( 3c \) into the \( A \) and \( B \) spots.

🎯 Exam Tip: When simplifying the squared terms, like \( \{2(a+b)\}^2 \), make sure to square both the coefficient (2) and the entire bracket \( (a+b) \).

Question 18. \( x^6 - 1 \)
Answer: We can factor \( x^6 - 1 \) in two ways. First, as a difference of squares: \( (x^3)^2 - (1)^2 \). Using the formula \( a^2 - b^2 = (a + b)(a - b) \), this becomes \( (x^3 + 1)(x^3 - 1) \). Now, both \( (x^3 + 1) \) and \( (x^3 - 1) \) are themselves sum and difference of cubes, respectively. For \( x^3 + 1 = (x)^3 + (1)^3 = (x + 1)(x^2 - x + 1) \). For \( x^3 - 1 = (x)^3 - (1)^3 = (x - 1)(x^2 + x + 1) \). Combining these, the complete factorization is \( (x + 1)(x^2 - x + 1)(x - 1)(x^2 + x + 1) \). You can also start by treating it as a difference of cubes, which might lead to the same result but requires more steps for the quadratic factors. This problem shows how different identities can be used sequentially.
\( x^6 - 1 = (x^3)^2 - (1)^2 \)
\( \implies (x^3 + 1)(x^3 - 1) \)
\( \implies [(x)^3 + (1)^3][(x)^3 - (1)^3] \)
\( \implies (x + 1)(x^2 - x + 1)(x - 1)(x^2 + x + 1) \)
\( \implies (x + 1)(x - 1)(x^2 - x + 1)(x^2 + x + 1) \) (reordered)
In simple words: First, think of \( x^6 - 1 \) as \( (x^3)^2 - (1)^2 \), which is a difference of squares. This gives you two brackets: \( (x^3 + 1) \) and \( (x^3 - 1) \). Then, factor each of these two brackets using the sum and difference of cubes rules.

🎯 Exam Tip: When you see an exponent of 6 (like \( x^6 \)), consider factoring it as both a difference of squares \( ((x^3)^2) \) and a difference of cubes \( ((x^2)^3) \). Starting with the difference of squares is usually simpler as it directly leads to the sum and difference of cubes expressions.

Question 19. \( 64a^6 - b^6 \)
Answer: Similar to the previous question, we can factor \( 64a^6 - b^6 \) as a difference of squares first: \( (8a^3)^2 - (b^3)^2 \). Using the formula \( X^2 - Y^2 = (X + Y)(X - Y) \), this becomes \( (8a^3 + b^3)(8a^3 - b^3) \). Now, both \( (8a^3 + b^3) \) and \( (8a^3 - b^3) \) are themselves sum and difference of cubes, respectively. For \( 8a^3 + b^3 = (2a)^3 + (b)^3 = (2a + b)(4a^2 - 2ab + b^2) \). For \( 8a^3 - b^3 = (2a)^3 - (b)^3 = (2a - b)(4a^2 + 2ab + b^2) \). Combining these, the complete factorization is \( (2a + b)(4a^2 - 2ab + b^2)(2a - b)(4a^2 + 2ab + b^2) \). We can rearrange the factors for a tidier appearance. This problem demonstrates the power of applying multiple factorization identities.
\( 64a^6 - b^6 = (8a^3)^2 - (b^3)^2 \)
\( \implies (8a^3 + b^3)(8a^3 - b^3) \)
\( \implies [(2a)^3 + (b)^3][(2a)^3 - (b)^3] \)
\( \implies (2a + b)( (2a)^2 - 2a \times b + b^2 )(2a - b)( (2a)^2 + 2a \times b + b^2 ) \)
\( \implies (2a + b)(4a^2 - 2ab + b^2)(2a - b)(4a^2 + 2ab + b^2) \)
\( \implies (2a + b)(2a - b)(4a^2 + 2ab + b^2)(4a^2 - 2ab + b^2) \) (reordered)
In simple words: First, see \( 64a^6 - b^6 \) as \( (8a^3)^2 - (b^3)^2 \), which is a difference of squares. This gives you two brackets. Then, factor each of these two brackets using the sum and difference of cubes rules.

🎯 Exam Tip: When the expressions become longer, keep track of all the factors. It's often helpful to write them down in a systematic order to avoid missing any terms.

Question 20. \( a^3 - b^3 + 4(a - b) \)
Answer: First, we know that \( a^3 - b^3 \) can be factored as \( (a - b)(a^2 + ab + b^2) \). So the expression becomes \( (a - b)(a^2 + ab + b^2) + 4(a - b) \). Now, we can see that \( (a - b) \) is a common factor in both terms. Factoring out \( (a - b) \), we get \( (a - b)((a^2 + ab + b^2) + 4) \), which simplifies to \( (a - b)(a^2 + ab + b^2 + 4) \). This factorization is a good example of identifying and extracting a common factor after applying an identity.
\( a^3 - b^3 + 4(a - b) \)
\( \implies (a - b)(a^2 + ab + b^2) + 4(a - b) \)
\( \implies (a - b)(a^2 + ab + b^2 + 4) \)
In simple words: First, break down \( a^3 - b^3 \) into \( (a-b)(a^2+ab+b^2) \). Then, notice that \( (a-b) \) is a common part in both terms. Take out \( (a-b) \) to get the final answer.

🎯 Exam Tip: Always look for common factors after applying a basic identity. This often leads to further simplification of the expression.

Question 21. \( x^3 - \frac{1}{x^3} - 6x + \frac{6}{x} \)
Answer: We can group the terms to simplify the expression. Group the cube terms and the terms with factor 6: \( (x^3 - \frac{1}{x^3}) - (6x - \frac{6}{x}) \). For the first group, \( x^3 - \frac{1}{x^3} \), which is \( (x)^3 - (\frac{1}{x})^3 \). Using the difference of cubes formula \( A^3 - B^3 = (A - B)(A^2 + AB + B^2) \), this becomes \( (x - \frac{1}{x})(x^2 + x \cdot \frac{1}{x} + (\frac{1}{x})^2) = (x - \frac{1}{x})(x^2 + 1 + \frac{1}{x^2}) \). For the second group, \( -6x + \frac{6}{x} = -6(x - \frac{1}{x}) \). Now, substitute these back into the original expression: \( (x - \frac{1}{x})(x^2 + 1 + \frac{1}{x^2}) - 6(x - \frac{1}{x}) \). We can see that \( (x - \frac{1}{x}) \) is a common factor. Factoring it out, we get \( (x - \frac{1}{x})((x^2 + 1 + \frac{1}{x^2}) - 6) \). This simplifies to \( (x - \frac{1}{x})(x^2 - 5 + \frac{1}{x^2}) \). Grouping terms and finding common factors are key strategies.
\( x^3 - \frac{1}{x^3} - 6x + \frac{6}{x} \)
\( \implies (x^3 - \frac{1}{x^3}) - (6x - \frac{6}{x}) \)
\( \implies (x - \frac{1}{x})(x^2 + 1 + \frac{1}{x^2}) - 6(x - \frac{1}{x}) \)
\( \implies (x - \frac{1}{x})(x^2 + 1 + \frac{1}{x^2} - 6) \)
\( \implies (x - \frac{1}{x})(x^2 - 5 + \frac{1}{x^2}) \)
In simple words: First, group the terms: \( (x^3 - \frac{1}{x^3}) \) and \( (-6x + \frac{6}{x}) \). Factor the first group using the difference of cubes. Factor \( -6 \) from the second group. You will see a common part \( (x - \frac{1}{x}) \), which you can then factor out.

🎯 Exam Tip: When factoring complex expressions, grouping terms strategically can reveal common factors or identities that were not obvious at first glance.

Question 22. \( 64a^3 + 125b^3 + 12a^2b + 15ab^2 \)
Answer: We can rewrite this expression by identifying the cubic terms and the remaining terms. \( 64a^3 = (4a)^3 \) and \( 125b^3 = (5b)^3 \). The other two terms are \( 12a^2b + 15ab^2 \). We can factor out a common term from these: \( 3ab(4a + 5b) \). So the entire expression becomes \( (4a)^3 + (5b)^3 + 3ab(4a + 5b) \). Now, apply the sum of cubes formula to \( (4a)^3 + (5b)^3 \): \( (4a + 5b)((4a)^2 - (4a)(5b) + (5b)^2) = (4a + 5b)(16a^2 - 20ab + 25b^2) \). Substituting this back, the expression is \( (4a + 5b)(16a^2 - 20ab + 25b^2) + 3ab(4a + 5b) \). Now, \( (4a + 5b) \) is a common factor. Factoring it out, we get \( (4a + 5b)((16a^2 - 20ab + 25b^2) + 3ab) \). This simplifies to \( (4a + 5b)(16a^2 - 17ab + 25b^2) \). This method combines grouping and identity application.
\( 64a^3 + 125b^3 + 12a^2b + 15ab^2 \)
\( \implies (4a)^3 + (5b)^3 + 3ab(4a + 5b) \)
\( \implies (4a + 5b)((4a)^2 - 4a \times 5b + (5b)^2) + 3ab(4a + 5b) \)
\( \implies (4a + 5b)(16a^2 - 20ab + 25b^2) + 3ab(4a + 5b) \)
\( \implies (4a + 5b)(16a^2 - 20ab + 25b^2 + 3ab) \)
\( \implies (4a + 5b)(16a^2 - 17ab + 25b^2) \)
In simple words: First, notice that \( 64a^3 \) is \( 4a \) cubed and \( 125b^3 \) is \( 5b \) cubed. Then, take out the common factor \( 3ab \) from \( 12a^2b + 15ab^2 \), which leaves \( (4a+5b) \). Now you have \( (4a)^3 + (5b)^3 + 3ab(4a+5b) \). Factor \( (4a)^3 + (5b)^3 \) using the sum of cubes rule. You will find \( (4a+5b) \) as a common factor in the entire expression, which you can take out.

🎯 Exam Tip: When an expression has four terms, consider if it fits the sum or difference of cubes formula with some common factoring. Look for perfect cubes, and then check if the remaining terms can be factored to reveal a common bracket.

Question 23. \( 375(a - b)^3 + 3 \)
Answer: First, we should always look for a common factor. We can factor out \( 3 \) from both terms: \( 3[125(a - b)^3 + 1] \). Now, the expression inside the bracket is \( 125(a - b)^3 + 1 \). This can be written as \( (5(a - b))^3 + (1)^3 \). This is a sum of cubes, using the formula \( X^3 + Y^3 = (X + Y)(X^2 - XY + Y^2) \). Let \( X = 5(a - b) = 5a - 5b \) and \( Y = 1 \). Applying the formula, the expression becomes \( 3[(5a - 5b + 1)((5a - 5b)^2 - (5a - 5b)(1) + (1)^2)] \). Now, expand \( (5a - 5b)^2 \): \( 25(a - b)^2 = 25(a^2 - 2ab + b^2) = 25a^2 - 50ab + 25b^2 \). So the full factorization is \( 3(5a - 5b + 1)(25a^2 + 25b^2 - 50ab - 5a + 5b + 1) \). Remember to simplify each part carefully.
\( 375(a - b)^3 + 3 \)
\( \implies 3[125(a - b)^3 + 1] \)
\( \implies 3[\{5(a - b)\}^3 + (1)^3] \)
\( \implies 3[\{5(a - b) + 1\}\{(5(a - b))^2 - 5(a - b) \times 1 + (1)^2\}] \)
\( \implies 3[(5a - 5b + 1)(25(a^2 - 2ab + b^2) - 5a + 5b + 1)] \)
\( \implies 3(5a - 5b + 1)(25a^2 + 25b^2 - 50ab - 5a + 5b + 1) \)
In simple words: First, take out the common number 3. Inside the bracket, notice that \( 125(a-b)^3 \) is \( (5(a-b))^3 \) and \( 1 \) is \( 1^3 \). Use the sum of cubes rule with \( X = 5(a-b) \) and \( Y = 1 \). Expand everything to get the final answer.

🎯 Exam Tip: Always look for common numerical factors first. This step makes the numbers smaller and easier to work with, especially when dealing with cubes.

Question 24. If \( a^4 + b^4 = a^2b^2 \), Show that \( a^6 + b^6 = 0 \)
Answer: We need to show that the Left Hand Side (L.H.S.) equals the Right Hand Side (R.H.S.), which is 0. Start with the L.H.S.: \( a^6 + b^6 \). We can rewrite \( a^6 + b^6 \) as a sum of cubes: \( (a^2)^3 + (b^2)^3 \). Using the sum of cubes formula \( X^3 + Y^3 = (X + Y)(X^2 - XY + Y^2) \), where \( X = a^2 \) and \( Y = b^2 \), this becomes: \( (a^2 + b^2)((a^2)^2 - a^2b^2 + (b^2)^2) \) \( = (a^2 + b^2)(a^4 - a^2b^2 + b^4) \) Now, we can rearrange the terms inside the second bracket: \( (a^2 + b^2)(a^4 + b^4 - a^2b^2) \). From the given condition, we know that \( a^4 + b^4 = a^2b^2 \). Substitute this into our expression: \( (a^2 + b^2)(a^2b^2 - a^2b^2) \). This simplifies to \( (a^2 + b^2)(0) \), which equals \( 0 \). Thus, L.H.S. = R.H.S. = 0. This problem showcases how to use a given condition to simplify and prove an identity.
Given: \( a^4 + b^4 = a^2b^2 \)
To Show: \( a^6 + b^6 = 0 \)
L.H.S. \( = a^6 + b^6 \)
\( \implies (a^2)^3 + (b^2)^3 \)
\( \implies (a^2 + b^2)((a^2)^2 - a^2b^2 + (b^2)^2) \)
\( \implies (a^2 + b^2)(a^4 - a^2b^2 + b^4) \)
\( \implies (a^2 + b^2)(a^4 + b^4 - a^2b^2) \)
Substitute \( a^4 + b^4 = a^2b^2 \) from the given condition:
\( \implies (a^2 + b^2)(a^2b^2 - a^2b^2) \)
\( \implies (a^2 + b^2) \times 0 \)
\( \implies 0 \)
\( \implies \text{R.H.S.} \)
Hence, \( a^6 + b^6 = 0 \).
In simple words: Start with \( a^6 + b^6 \). Change it to \( (a^2)^3 + (b^2)^3 \). Use the sum of cubes formula. Then, use the information given in the question \( (a^4 + b^4 = a^2b^2) \) to replace a part of your answer. This will make the expression zero.

🎯 Exam Tip: In "show that" or "prove that" questions, clearly state your L.H.S and R.H.S. and use the given conditions logically to transform one side into the other. This structured approach helps ensure all steps are clear and correct.