Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.4

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.4

Question 1. Expand the following:
(i) \( (2x + 3y + 4z)^2 \)
(ii) \( (-p + 2q + 3r)^2 \)
(iii) \( (2p + 3) (2p - 4) (2p - 5) \)
(iv) \( (3a + 1) (3a - 2) (3a + 4) \)
Answer:
(i) We use the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \).
So, for \( (2x + 3y + 4z)^2 \):
\( (2x)^2 + (3y)^2 + (4z)^2 + 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x) \)
\( = 4x^2 + 9y^2 + 16z^2 + 12xy + 24yz + 16xz \)

(ii) We use the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac \).
So, for \( (-p + 2q + 3r)^2 \):
\( (-p)^2 + (2q)^2 + (3r)^2 + 2(-p)(2q) + 2(2q)(3r) + 2(3r)(-p) \)
\( = p^2 + 4q^2 + 9r^2 - 4pq + 12qr - 6pr \)

(iii) We use the identity \( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ac)x + abc \).
For \( (2p + 3) (2p - 4) (2p - 5) \), we have \( x = 2p \), \( a = 3 \), \( b = -4 \), and \( c = -5 \).
Substitute these values into the identity:
\( (2p)^3 + (3 - 4 - 5)(2p)^2 + [(3)(-4) + (-4)(-5) + (3)(-5)](2p) + (3)(-4)(-5) \)
\( = 8p^3 + (-6)(4p^2) + [-12 + 20 - 15](2p) + 60 \)
\( = 8p^3 - 24p^2 + (-7)(2p) + 60 \)
\( = 8p^3 - 24p^2 - 14p + 60 \)

(iv) We use the identity \( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ac)x + abc \).
For \( (3a + 1) (3a - 2) (3a + 4) \), we have \( x = 3a \), \( a = 1 \), \( b = -2 \), and \( c = 4 \).
Substitute these values into the identity:
\( (3a)^3 + (1 - 2 + 4)(3a)^2 + [(1)(-2) + (-2)(4) + (1)(4)](3a) + (1)(-2)(4) \)
\( = 27a^3 + (3)(9a^2) + [-2 - 8 + 4](3a) - 8 \)
\( = 27a^3 + 27a^2 + (-6)(3a) - 8 \)
\( = 27a^3 + 27a^2 - 18a - 8 \)
In simple words: To expand these expressions, we use specific algebraic formulas. For three terms squared, we square each term and add twice the product of each pair. For three terms multiplied, we use a formula involving the sum of the numbers, the sum of their products in pairs, and the product of all three numbers.

🎯 Exam Tip: Remember the basic algebraic identities like \( (a+b+c)^2 \) and \( (x+a)(x+b)(x+c) \) and practice applying them carefully, especially with negative signs.

Question 2. Using algebraic identity, find the coefficients of \( x^2 \), \( x \) and constant term without actual expansion.
(i) \( (x + 5)(x + 6)(x + 7) \)
(ii) \( (2x + 3)(2x - 5) (2x - 6) \)
Answer:
We use the identity \( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ac)x + abc \).

(i) For \( (x + 5)(x + 6)(x + 7) \):
Here, the variable is \( x \), and \( a = 5 \), \( b = 6 \), \( c = 7 \).
Coefficient of \( x^2 = a + b + c = 5 + 6 + 7 = 18 \)
Coefficient of \( x = ab + bc + ac = (5)(6) + (6)(7) + (5)(7) = 30 + 42 + 35 = 107 \)
Constant term \( = abc = (5)(6)(7) = 210 \)

(ii) For \( (2x + 3)(2x - 5) (2x - 6) \):
Here, the variable part is \( 2x \), and \( a = 3 \), \( b = -5 \), \( c = -6 \).
When we compare \( (2x + a)(2x + b)(2x + c) \) with the identity, the \( x^2 \) term in the identity \( (a+b+c)x^2 \) becomes \( (a+b+c)(2x)^2 \).
Coefficient of \( x^2 = (a + b + c) \times (\text{coefficient of } x)^2 \)
\( = (3 - 5 - 6) \times (2)^2 \)
\( = (-8) \times 4 \)
\( = -32 \)
Similarly, the \( x \) term in the identity \( (ab + bc + ac)x \) becomes \( (ab + bc + ac)(2x) \).
Coefficient of \( x = (ab + bc + ac) \times (\text{coefficient of } x) \)
\( = [(3)(-5) + (-5)(-6) + (-6)(3)] \times 2 \)
\( = [-15 + 30 - 18] \times 2 \)
\( = [-3] \times 2 \)
\( = -6 \)
Constant term \( = abc = (3)(-5)(-6) \)
\( = 90 \)
In simple words: We can find the numbers in front of \( x^2 \) and \( x \), and the number alone (constant term) by using a special algebraic rule. For part (ii), because \( x \) has a number (2) in front of it, we need to multiply the coefficients by that number (or its square) too.

🎯 Exam Tip: When the variable term is not just \( x \) (e.g., \( 2x \)), remember to adjust the coefficients of \( x^2 \) and \( x \) by multiplying by the square of the coefficient of \( x \) and the coefficient of \( x \) respectively.

Question 3. If \( (x + a)(x + b)(x + c) = x^3 + 14x^2 + 59x + 70 \), find the value of
(i) \( a + b + c \)
(ii) \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \)
(iii) \( a^2 + b^2 + c^2 \)
(iv) \( \frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab} \)
Answer:
We know the identity: \( (x + a)(x + b)(x + c) = x^3 + (a + b + c)x^2 + (ab + bc + ac)x + abc \).
Comparing this with the given equation \( x^3 + 14x^2 + 59x + 70 \), we get:
\( a + b + c = 14 \)
\( ab + bc + ac = 59 \)
\( abc = 70 \)

(i) From the comparison, \( a + b + c = 14 \).

(ii) To find \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \), we can find a common denominator:
\( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{bc}{abc} + \frac{ac}{abc} + \frac{ab}{abc} = \frac{ab + bc + ac}{abc} \)
Substitute the values we found:
\( = \frac{59}{70} \)

(iii) We use the identity \( a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ac) \).
Substitute the values:
\( = (14)^2 - 2(59) \)
\( = 196 - 118 \)
\( = 78 \)

(iv) To find \( \frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab} \), we find a common denominator:
\( \frac{a}{bc} + \frac{b}{ac} + \frac{c}{ab} = \frac{a \times a}{bc \times a} + \frac{b \times b}{ac \times b} + \frac{c \times c}{ab \times c} = \frac{a^2 + b^2 + c^2}{abc} \)
Substitute the values we found in part (iii) and from the comparison:
\( = \frac{78}{70} \)
We can simplify this fraction by dividing both the numerator and denominator by 2:
\( = \frac{39}{35} \)
In simple words: By comparing the given polynomial with its standard expanded form, we can quickly find the values of \( a+b+c \), \( ab+bc+ac \), and \( abc \). Then, we use simple fraction rules and another identity to find the other values.

🎯 Exam Tip: Memorize the expansions of \( (x+a)(x+b)(x+c) \) and the identity for \( a^2+b^2+c^2 \) as they are commonly used in such problems. Look for opportunities to simplify fractions in your final answer.

Question 4. Expand:
(i) \( (3a - 4b)^3 \)
(ii) \( (x + \frac{1}{y})^3 \)
Answer:
(i) We use the identity \( (A - B)^3 = A^3 - B^3 - 3AB(A - B) \).
For \( (3a - 4b)^3 \), we have \( A = 3a \) and \( B = 4b \).
Substitute these values:
\( (3a)^3 - (4b)^3 - 3(3a)(4b)(3a - 4b) \)
\( = 27a^3 - 64b^3 - 36ab(3a - 4b) \)
Now, distribute \( -36ab \) into the parenthesis:
\( = 27a^3 - 64b^3 - (36ab)(3a) + (36ab)(4b) \)
\( = 27a^3 - 64b^3 - 108a^2b + 144ab^2 \)

(ii) We use the identity \( (A + B)^3 = A^3 + B^3 + 3AB(A + B) \).
For \( (x + \frac{1}{y})^3 \), we have \( A = x \) and \( B = \frac{1}{y} \).
Substitute these values:
\( x^3 + (\frac{1}{y})^3 + 3(x)(\frac{1}{y})(x + \frac{1}{y}) \)
\( = x^3 + \frac{1}{y^3} + \frac{3x}{y}(x + \frac{1}{y}) \)
Now, distribute \( \frac{3x}{y} \) into the parenthesis:
\( = x^3 + \frac{1}{y^3} + \frac{3x^2}{y} + \frac{3x}{y^2} \)
In simple words: To expand expressions raised to the power of 3, we use special cube formulas. For subtraction, it's \( (A-B)^3 \), and for addition, it's \( (A+B)^3 \). We carefully put the values into the formula and then multiply everything out.

🎯 Exam Tip: Be very careful with signs when using the \( (A-B)^3 \) identity. Also, remember to distribute terms correctly after applying the main identity.

Question 5. Evaluate the following by using identities:
(i) \( 98^3 \)
(ii) \( 1001^3 \)
Answer:
(i) We want to evaluate \( 98^3 \). We can write 98 as \( 100 - 2 \).
So, we use the identity \( (A - B)^3 = A^3 - B^3 - 3AB(A - B) \).
Here, \( A = 100 \) and \( B = 2 \).
\( (100 - 2)^3 = (100)^3 - (2)^3 - 3(100)(2)(100 - 2) \)
\( = 1000000 - 8 - 600(98) \)
\( = 1000000 - 8 - 58800 \)
\( = 999992 - 58800 \)
\( = 941192 \)

(ii) We want to evaluate \( 1001^3 \). We can write 1001 as \( 1000 + 1 \).
So, we use the identity \( (A + B)^3 = A^3 + B^3 + 3AB(A + B) \).
Here, \( A = 1000 \) and \( B = 1 \).
\( (1000 + 1)^3 = (1000)^3 + (1)^3 + 3(1000)(1)(1000 + 1) \)
\( = 1000000000 + 1 + 3000(1001) \)
\( = 1000000001 + 3003000 \)
\( = 1003003001 \)
In simple words: To calculate large numbers cubed easily, we change them into sums or differences of easier numbers like 100 or 1000. Then we use the cube formulas to quickly find the answer without doing long multiplications.

🎯 Exam Tip: Choose numbers that make calculations simple (like 100-2 or 1000+1). This avoids complex arithmetic and reduces the chance of errors. Clearly show each step of the identity application.

Question 6. If \( (x + y + z) = 9 \) and \( (xy + yz + zx) = 26 \), then find the value of \( x^2 + y^2 + z^2 \).
Answer:
We are given:
\( x + y + z = 9 \)
\( xy + yz + zx = 26 \)
We know the identity: \( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \).
To find \( x^2 + y^2 + z^2 \), we can rearrange the identity:
\( x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + zx) \)
Now, substitute the given values into the rearranged identity:
\( x^2 + y^2 + z^2 = (9)^2 - 2(26) \)
\( = 81 - 52 \)
\( = 29 \)
In simple words: We have a special math rule that connects the sum of three numbers, the sum of their squares, and the sum of their products in pairs. By using this rule and the numbers we are given, we can find the sum of their squares.

🎯 Exam Tip: This problem directly tests the identity \( (x+y+z)^2 \). Make sure you know this formula and how to rearrange it to find any of its components if the others are given.

Question 7. Find \( 27a^3 + 64b^3 \), if \( 3a + 4b = 10 \) and \( ab = 2 \).
Answer:
We are given:
\( 3a + 4b = 10 \)
\( ab = 2 \)
We need to find \( 27a^3 + 64b^3 \). We can write this as \( (3a)^3 + (4b)^3 \).
We use the identity: \( A^3 + B^3 = (A + B)^3 - 3AB(A + B) \).
Here, \( A = 3a \) and \( B = 4b \).
Substitute these into the identity:
\( (3a)^3 + (4b)^3 = (3a + 4b)^3 - 3(3a)(4b)(3a + 4b) \)
\( = (3a + 4b)^3 - 36ab(3a + 4b) \)
Now, substitute the given values \( (3a + 4b = 10) \) and \( (ab = 2) \):
\( = (10)^3 - 36(2)(10) \)
\( = 1000 - 720 \)
\( = 280 \)
In simple words: We want to find the sum of two cube terms. We use a special formula that connects this sum to the sum of the base terms and their product. We plug in the numbers we already know to get the answer.

🎯 Exam Tip: Recognize that \( 27a^3 \) is \( (3a)^3 \) and \( 64b^3 \) is \( (4b)^3 \). This helps in applying the sum of cubes identity \( A^3+B^3 = (A+B)^3 - 3AB(A+B) \) correctly.

Question 8. Find \( x^3 - y^3 \), if \( x - y = 5 \) and \( xy = 14 \).
Answer:
We are given:
\( x - y = 5 \)
\( xy = 14 \)
We need to find \( x^3 - y^3 \).
We use the identity: \( A^3 - B^3 = (A - B)^3 + 3AB(A - B) \).
Here, \( A = x \) and \( B = y \).
Substitute the given values into the identity:
\( x^3 - y^3 = (x - y)^3 + 3xy(x - y) \)
\( = (5)^3 + 3(14)(5) \)
\( = 125 + 210 \)
\( = 335 \)
In simple words: We want to find the difference between two cube terms. We use a special formula that connects this difference to the difference of the base terms and their product. We just put in the numbers we already know to calculate the final answer.

🎯 Exam Tip: Be careful with the signs when using the difference of cubes identity: \( A^3 - B^3 = (A - B)^3 + 3AB(A - B) \). It's easy to confuse it with the sum of cubes identity or other similar formulas.

Question 9. If \( a + \frac{1}{a} = 6 \), then find the value of \( a^3 + \frac{1}{a^3} \).
Answer:
We are given: \( a + \frac{1}{a} = 6 \).
We need to find \( a^3 + \frac{1}{a^3} \).
We use the identity: \( A^3 + B^3 = (A + B)^3 - 3AB(A + B) \).
Here, \( A = a \) and \( B = \frac{1}{a} \).
Substitute these into the identity:
\( a^3 + (\frac{1}{a})^3 = (a + \frac{1}{a})^3 - 3(a)(\frac{1}{a})(a + \frac{1}{a}) \)
Notice that \( a \times \frac{1}{a} = 1 \). So the expression simplifies to:
\( = (a + \frac{1}{a})^3 - 3(1)(a + \frac{1}{a}) \)
\( = (a + \frac{1}{a})^3 - 3(a + \frac{1}{a}) \)
Now, substitute the given value \( (a + \frac{1}{a} = 6) \):
\( = (6)^3 - 3(6) \)
\( = 216 - 18 \)
\( = 198 \)
In simple words: When we know the sum of a number and its inverse, we can find the sum of their cubes using a special formula. The product of the number and its inverse is always 1, which makes the calculation simpler.

🎯 Exam Tip: For expressions involving a term and its reciprocal (like \( a \) and \( \frac{1}{a} \)), remember that their product is 1. This simplifies the \( 3AB(A+B) \) part of the cube identity to \( 3(A+B) \).

Question 10. If \( x^2 + \frac{1}{x^2} = 23 \), then find the value of \( x + \frac{1}{x} \) and \( x^3 + \frac{1}{x^3} \).
Answer:
We are given: \( x^2 + \frac{1}{x^2} = 23 \).
**Part 1: Find \( x + \frac{1}{x} \)**
We know the identity \( (A + B)^2 = A^2 + B^2 + 2AB \).
Let \( A = x \) and \( B = \frac{1}{x} \). Then \( (x + \frac{1}{x})^2 = x^2 + (\frac{1}{x})^2 + 2(x)(\frac{1}{x}) \).
Since \( x \times \frac{1}{x} = 1 \), this simplifies to:
\( (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 \)
Substitute the given value \( x^2 + \frac{1}{x^2} = 23 \):
\( (x + \frac{1}{x})^2 = 23 + 2 \)
\( (x + \frac{1}{x})^2 = 25 \)
Take the square root of both sides:
\( x + \frac{1}{x} = \sqrt{25} \)
\( x + \frac{1}{x} = \pm 5 \)

**Part 2: Find \( x^3 + \frac{1}{x^3} \)**
We use the identity: \( A^3 + B^3 = (A + B)^3 - 3AB(A + B) \).
Here, \( A = x \) and \( B = \frac{1}{x} \). As shown in Question 9, this simplifies to:
\( x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) \)
We have two possible values for \( x + \frac{1}{x} \), which are \( 5 \) and \( -5 \). We calculate for both:
**Case 1: If \( x + \frac{1}{x} = 5 \)**
\( x^3 + \frac{1}{x^3} = (5)^3 - 3(5) \)
\( = 125 - 15 \)
\( = 110 \)
**Case 2: If \( x + \frac{1}{x} = -5 \)**
\( x^3 + \frac{1}{x^3} = (-5)^3 - 3(-5) \)
\( = -125 + 15 \)
\( = -110 \)
So, \( x^3 + \frac{1}{x^3} = \pm 110 \).
In simple words: First, we use the square of a sum identity to find the value of \( x + \frac{1}{x} \). Since it's a square root, there will be two possible answers (positive and negative). Then, we use the sum of cubes identity, applying it for both positive and negative results from the first step to get the final answers.

🎯 Exam Tip: When taking the square root, always remember to include both positive and negative solutions (\( \pm \)). Also, ensure you use the correct value of \( (x + \frac{1}{x}) \) in the cube identity for each case.

Question 11. If \( (y - \frac{1}{y})^3 = 27 \) then find the value of \( y^3 - \frac{1}{y^3} \).
Answer:
We are given: \( (y - \frac{1}{y})^3 = 27 \).
To find \( y - \frac{1}{y} \), we take the cube root of both sides:
\( \sqrt[3]{(y - \frac{1}{y})^3} = \sqrt[3]{27} \)
\( y - \frac{1}{y} = 3 \)
Now we need to find \( y^3 - \frac{1}{y^3} \).
We use the identity: \( A^3 - B^3 = (A - B)^3 + 3AB(A - B) \).
Here, \( A = y \) and \( B = \frac{1}{y} \).
Since \( y \times \frac{1}{y} = 1 \), the expression simplifies to:
\( y^3 - \frac{1}{y^3} = (y - \frac{1}{y})^3 + 3(y - \frac{1}{y}) \)
Substitute the value \( (y - \frac{1}{y} = 3) \) that we found:
\( y^3 - \frac{1}{y^3} = (3)^3 + 3(3) \)
\( = 27 + 9 \)
\( = 36 \)
In simple words: First, we find the value of \( y - \frac{1}{y} \) by taking the cube root of the given equation. Then, we use a special formula for the difference of cubes. This formula helps us find \( y^3 - \frac{1}{y^3} \) by using the value of \( y - \frac{1}{y} \).

🎯 Exam Tip: Remember that taking a cube root gives only one real solution, unlike a square root which gives positive and negative solutions. Clearly identify \( (A-B) \) and \( AB \) for the identity.

Question 12. Simplify:
(i) \( (2a + 3b + 4c) (4a^2 + 9b^2 + 16c^2 - 6ab - 12bc - 8ca) \)
(ii) \( (x - 2y + 3z) (x^2 + 4y^2 + 9z^2 + 2xy + 6yz - 3xz) \)
Answer:
We use the identity: \( A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 + B^2 + C^2 - AB - BC - CA) \).

(i) Compare the given expression with the right side of the identity:
\( (2a + 3b + 4c) ( (2a)^2 + (3b)^2 + (4c)^2 - (2a)(3b) - (3b)(4c) - (4c)(2a) ) \)
Here, \( A = 2a \), \( B = 3b \), \( C = 4c \).
So, the simplified form is \( A^3 + B^3 + C^3 - 3ABC \):
\( (2a)^3 + (3b)^3 + (4c)^3 - 3(2a)(3b)(4c) \)
\( = 8a^3 + 27b^3 + 64c^3 - 72abc \)

(ii) Compare the given expression with the right side of the identity:
\( (x - 2y + 3z) ( x^2 + (-2y)^2 + (3z)^2 - (x)(-2y) - (-2y)(3z) - (3z)(x) ) \)
Here, \( A = x \), \( B = -2y \), \( C = 3z \).
So, the simplified form is \( A^3 + B^3 + C^3 - 3ABC \):
\( (x)^3 + (-2y)^3 + (3z)^3 - 3(x)(-2y)(3z) \)
\( = x^3 - 8y^3 + 27z^3 - (-18xyz) \)
\( = x^3 - 8y^3 + 27z^3 + 18xyz \)
In simple words: We use a special algebra rule that helps to simplify a product of two big expressions into a sum of cubes minus three times the product of the terms. We just need to identify what each 'A', 'B', and 'C' stands for, making sure to include any negative signs.

🎯 Exam Tip: The key to these problems is recognizing the expanded form of \( A^3+B^3+C^3-3ABC \). Pay close attention to the signs in the given expression to correctly identify \( A, B, \) and \( C \).

Question 13. By using identity evaluate the following:
(i) \( 7^3 - 10^3 + 3^3 \)
(ii) \( 1 + \frac{1}{8} - \frac{27}{8} \)
Answer:
We use the special identity: If \( A + B + C = 0 \), then \( A^3 + B^3 + C^3 = 3ABC \).

(i) We need to evaluate \( 7^3 - 10^3 + 3^3 \).
Let \( A = 7 \), \( B = -10 \), \( C = 3 \).
Check if \( A + B + C = 0 \):
\( 7 + (-10) + 3 = 7 - 10 + 3 = -3 + 3 = 0 \)
Since \( A + B + C = 0 \), we can use the identity:
\( A^3 + B^3 + C^3 = 3ABC \)
\( 7^3 + (-10)^3 + 3^3 = 3(7)(-10)(3) \)
\( = 21 \times (-30) \)
\( = -630 \)

(ii) We need to evaluate \( 1 + \frac{1}{8} - \frac{27}{8} \).
We can write these terms as cubes:
\( 1 = 1^3 \)
\( \frac{1}{8} = (\frac{1}{2})^3 \)
\( -\frac{27}{8} = (-\frac{3}{2})^3 \)
So, let \( A = 1 \), \( B = \frac{1}{2} \), \( C = -\frac{3}{2} \).
Check if \( A + B + C = 0 \):
\( 1 + \frac{1}{2} - \frac{3}{2} = \frac{2}{2} + \frac{1}{2} - \frac{3}{2} = \frac{2 + 1 - 3}{2} = \frac{0}{2} = 0 \)
Since \( A + B + C = 0 \), we can use the identity:
\( A^3 + B^3 + C^3 = 3ABC \)
\( 1^3 + (\frac{1}{2})^3 + (-\frac{3}{2})^3 = 3(1)(\frac{1}{2})(-\frac{3}{2}) \)
\( = 3 \times \frac{1}{2} \times (-\frac{3}{2}) \)
\( = -\frac{9}{4} \)
In simple words: When the sum of three numbers is zero, their individual cubes added together equal three times their product. We check if the numbers in the problem add up to zero, then apply this useful shortcut to find the answer quickly.

🎯 Exam Tip: Always check if \( A+B+C=0 \) before applying the identity \( A^3+B^3+C^3=3ABC \). This identity is a powerful shortcut but only works under that specific condition. Convert fractions to common denominators when checking the sum.

Question 14. If \( 2x - 3y - 4z = 0 \), then find \( 8x^3 - 27y^3 - 64z^3 \).
Answer:
We are given: \( 2x - 3y - 4z = 0 \).
We need to find \( 8x^3 - 27y^3 - 64z^3 \).
We can rewrite this expression with cube terms:
\( (2x)^3 + (-3y)^3 + (-4z)^3 \)
Let \( A = 2x \), \( B = -3y \), \( C = -4z \).
From the given condition, \( A + B + C = 2x + (-3y) + (-4z) = 2x - 3y - 4z = 0 \).
Since \( A + B + C = 0 \), we can use the identity: \( A^3 + B^3 + C^3 = 3ABC \).
Substitute \( A, B, C \) back into the identity:
\( (2x)^3 + (-3y)^3 + (-4z)^3 = 3(2x)(-3y)(-4z) \)
\( = 3 \times 2x \times (-3y) \times (-4z) \)
\( = 6x \times 12yz \)
\( = 72xyz \)
So, \( 8x^3 - 27y^3 - 64z^3 = 72xyz \).
In simple words: If three terms add up to zero, then the sum of their individual cubes is equal to three times their product. We first identify the three terms (making sure to keep their signs) and confirm they add to zero. Then, we simply multiply them together and by 3 to get the answer.

🎯 Exam Tip: When given an expression like \( 2x - 3y - 4z = 0 \), be sure to define your A, B, and C terms correctly, including their signs, before applying the identity \( A^3+B^3+C^3=3ABC \).