Algebraic Expressions

Q1: Multiple Choice Type

i. \(\left(-18xy\right)-(-8xy)\) is equal to:

Step 1: Use the identity: \(a – (-b) = a + b\)
\(-18xy – (-8xy) = -18xy + 8xy\)
Step 2: Combine like terms:
\(-18xy + 8xy = -10xy\)
Answer: b. \(-10xy\)

ii. \((9a + 7b – 6c) – (2a – 3b + 4c)\) is equal to:

Step 1: Distribute the minus sign:
\(9a + 7b – 6c – 2a + 3b – 4c\)
Step 2: Combine like terms:
\(9a – 2a = 7a\)
\(7b + 3b = 10b\)
\(-6c – 4c = -10c\)
Final Expression:
\(7a + 10b – 10c\)
Answer: b. \(7a + 10b – 10c\)

iii. \(-81a^5b^4c^3 \div (-9a^2b^2c)\) is equal to:

Step 1: Divide coefficients:
\(-81 \div -9 = 9\)
Step 2: Apply exponent division:
\(a^5 \div a^2 = a^{5-2} = a^3\)
\(b^4 \div b^2 = b^2\)
\(c^3 \div c = c^2\)
Final Result:
\(9a^3b^2c^2\)
Answer: d. \(9a^3b^2c^2\)

iv. \(-\left(-pq – \overline{p^2 – pq}\right)\) is equal to:

Step 1: Simplify inside the bar:
\(\overline{p^2 – pq} = p^2 – pq\)
Step 2: Insert into expression:
\(-(-pq – (p^2 – pq)) = -(-pq – p^2 + pq)\)
Step 3: Simplify inside:
\(-pq + pq = 0 \Rightarrow -p^2\)
\(-(-p^2) = p^2\)
Answer: c. \(p^2\)

v. \(x^3 – y^3 – x(y^2 + \overline{x^2 – z^2})\) is equal to:

Step 1: Simplify the bar:
\(\overline{x^2 – z^2} = x^2 – z^2\)
Step 2: Combine terms:
\(x^3 – y^3 – x(y^2 + x^2 – z^2)\)
\(x^3 – y^3 – x(y^2 + x^2 – z^2)\)
Step 3: Multiply:
\(x^3 – y^3 – (x y^2 + x^3 – x z^2)\)
Step 4: Combine like terms:
\(x^3 – y^3 – x^3 – xy^2 + xz^2\)
Cancel \(x^3\):
\(-y^3 – xy^2 + xz^2\)
Answer: d. \(-y^3 – xy^2 + xz^2\)

vi. Statement 1: The expression \(2x^4 – 3x^2 + \frac{7}{x},\ x \ne 0\) has no constant term.
Statement 2: In an algebraic expression in terms of one variable, the term(s) independent of the variable is called the constant.

Step 1: Statement 1 analysis:
Expression: \(2x^4 – 3x^2 + \frac{7}{x}\)
All terms contain x either as positive or negative powers. No constant term.
✅ So, Statement 1 is TRUE.
Step 2: Statement 2 is a correct definition of a constant term.
✅ So, Statement 2 is also TRUE.
Answer: a. Both the statements are true.

vii. Assertion (A): \(5x + y^2 – x^3\), \(xy + yz + zx\), \(x^2 – x + 1\) are all trinomials.
Reason (R): An algebraic expression which contains three different terms is called a trinomial.

Step 1: Check Assertion (A)
– \(5x + y^2 – x^3\) → 3 terms ✔️
– \(xy + yz + zx\) → 3 terms ✔️
– \(x^2 – x + 1\) → 3 terms ✔️
✅ A is true.
Step 2: Check Reason (R)
The definition given is correct: a trinomial has exactly three distinct terms.
✅ R is true.
Conclusion: Both A and R are true, and R explains A.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

viii. Assertion (A): Refer to the following algebraic expressions in terms of one variable: \(3x+9,\ 9-\frac{3}{x^2},\ 100\). Two of them are not x.
Reason (R): An algebraic expression is a polynomial if the power of each term used in it is a whole number.

Step 1: Check each expression:
– \(3x + 9\) → polynomial in x ✔️
– \(9 – \frac{3}{x^2}\) → contains \(x^{-2}\), power not a whole number ❌
– \(100\) → constant polynomial ✔️
✅ Only one is in x: Assertion A is false.
Step 2: Reason (R) is correct definition of a polynomial.
Answer: d. A is false, but R is true.

ix. Assertion (A): \(2xyz + 3x^2y\) is a cubic polynomial in three variables.
Reason (R): In algebraic expressions with multiple variables, the degree is the sum of powers in a term, if this sum is a whole number.

Step 1: Find degree of each term:
– \(2xyz\) → degree = 1+1+1 = 3
– \(3x^2y\) → degree = 2+1 = 3
✅ Highest degree = 3 → A is true ✔️
Step 2: R is the correct rule for degree in multivariable expressions.
✅ R is true ✔️
Conclusion: Both A and R are correct, and R explains A.
Answer: a. Both A and R are correct, and R is the correct explanation for A.

x. Assertion (A): \(2024x^2yz\) is not a trinomial.
Reason (R): For polynomials, the like terms of the given polynomials are combined together.

Step 1: Understand the definition of a trinomial.
A trinomial is a polynomial expression that has exactly three unlike terms.
e.g., \(3x + 4y – 5\) is a trinomial.
Step 2: Analyze the expression \(2024x^2yz\)
This is a single term — only one monomial is present.
So, it is not a trinomial. ✅ Assertion is correct.
Step 3: Analyze the Reason (R)
The Reason says: “For polynomials, like terms are combined together.”
This is a true statement, but it has nothing to do with the number of terms in \(2024x^2yz\), because it’s already a monomial.
Step 4: Conclusion
The Assertion (A) is true, and Reason (R) is also true, but R is not the correct explanation for A.
Answer: b. Both A and R are correct, and R is the explanation for A.

Q2: Given the term \(\frac{5}{7}xy^2z^3\), find the coefficient of:

i. Coefficient of 5

Step 1: The term is \(\frac{5}{7}xy^2z^3\)
Step 2: 5 is the numerator of the coefficient.
Answer: \(\frac{1}{7}xy^2z^3\)

ii. Coefficient of \(\frac{5}{7}\)

Answer: \(xy^2z^3\)

iii. Coefficient of 5x

Answer: \(\frac{1}{7}y^2z^3\)

iv. Coefficient of \(xy^2\)

Answer: \(\frac{5}{7}z^3\)

v. Coefficient of \(z^3\)

Answer: \(\frac{5}{7}xy^2\)

vi. Coefficient of \(xz^3\)

Answer: \(\frac{5}{7}y^2\)

vii. Coefficient of \(5xy^2\)

Answer: \(\frac{1}{7}z^3\)

viii. Coefficient of \(\frac{1}{7}yz\)

Answer: \(5yz^2x\)

ix. Coefficient of \(z\)

Answer: \(\frac{5}{7}xy^2z^2\)

x. Coefficient of \(yz^2\)

Answer: \(\frac{5}{7}xyz\)

xi. Coefficient of \(5xyz\)

Answer: \(\frac{1}{7}yz^2\)

Q3: In each polynomial given below, separate the like terms:

i. \(3xy,\ -4yx^2,\ 2xy^2,\ 2.5x^2y,\ -8yx,\ -3.2y^2x,\ x^2y\)

Step 1: Rearranging the terms to standard format:
\(\Rightarrow 3xy,\ -8xy,\ 2xy^2,\ -3.2xy^2,\ -4x^2y,\ 2.5x^2y,\ x^2y\)
Step 2: Grouping like terms:
– Like terms for \(xy\): \(3xy,\ -8xy\)
– Like terms for \(xy^2\): \(2xy^2,\ -3.2xy^2\)
– Like terms for \(x^2y\): \(-4x^2y,\ 2.5x^2y,\ x^2y\)
Answer:
Like terms:
→ \(3xy,\ -8xy\)
→ \(2xy^2,\ -3.2xy^2\)
→ \(-4x^2y,\ 2.5x^2y,\ x^2y\)

ii. \(y^2z^3,\ xy^2z^3,\ -5x^2yz,\ -4y^2z^3,\ -8xz^3y^2,\ 3x^2yz,\ 2z^3y^2\)

Step 1: Rearranging the terms to standard format:
\(\Rightarrow y^2z^3,\ -4y^2z^3,\ 2y^2z^3,\ xy^2z^3,\ -8xy^2z^3,\ -5x^2yz,\ 3x^2yz\)
Step 2: Grouping like terms:
– Like terms for \(y^2z^3\): \(y^2z^3,\ -4y^2z^3,\ 2z^3y^2\) (All same)
– Like terms for \(xy^2z^3\): \(xy^2z^3,\ -8xz^3y^2\) (same variables, same powers)
– Like terms for \(x^2yz\): \(-5x^2yz,\ 3x^2yz\)
Answer:
Like terms:
→ \(y^2z^3,\ -4y^2z^3,\ 2z^3y^2\)
→ \(xy^2z^3,\ -8xz^3y^2\)
→ \(-5x^2yz,\ 3x^2yz\)

Q4: The sides of a triangle are \(x^2 – 3xy + 8\), \(4x^2 + 5xy – 3\) and \(6 – 3x^2 + 4xy\). Find its perimeter.

Step 1: Write the expressions for the three sides of the triangle.
Side 1 = \(x^2 – 3xy + 8\)
Side 2 = \(4x^2 + 5xy – 3\)
Side 3 = \(6 – 3x^2 + 4xy\)
Step 2: Perimeter of a triangle = Sum of its three sides
⇒ Perimeter = \((x^2 – 3xy + 8) + (4x^2 + 5xy – 3) + (6 – 3x^2 + 4xy)\)
Step 3: Remove brackets and combine all like terms
= \(x^2 + 4x^2 – 3x^2\)
+ \(-3xy + 5xy + 4xy\)
+ \(8 – 3 + 6\)
Step 4: Add the like terms
= \((1 + 4 – 3)x^2 = 2x^2\)
= \((-3 + 5 + 4)xy = 6xy\)
= \(8 – 3 + 6 = 11\)
Answer: The perimeter of the triangle is \(2x^2 + 6xy + 11\)

Q5: The perimeter of a triangle is \(8y^2 – 9y + 4\) and its two sides are \(3y^2 – 5y\) and \(4y^2 + 12\). Find its third side.

Step 1: Let the third side be \(x\).
Perimeter = Sum of the three sides
⇒ \(8y^2 – 9y + 4 = (3y^2 – 5y) + (4y^2 + 12) + x\)
Step 2: Add the two known sides:
\((3y^2 – 5y) + (4y^2 + 12) = 3y^2 + 4y^2 – 5y + 12 = 7y^2 – 5y + 12\)
Step 3: Now subtract this sum from the perimeter to find the third side: \[ x = (8y^2 – 9y + 4) – (7y^2 – 5y + 12) \]Step 4: Subtract term-by-term: \[ x = 8y^2 – 9y + 4 – 7y^2 + 5y – 12 \\ x = (8y^2 – 7y^2) + (-9y + 5y) + (4 – 12) \\ x = y^2 – 4y – 8 \]Answer: The third side is \(y^2 – 4y – 8\)

Q6: The two adjacent sides of a rectangle are \(2x^2 – 5xy + 3z^2\) and \(4xy – x^2 – z^2\). Find its perimeter.

Step 1: Perimeter of a rectangle = 2 × (Length + Breadth)
Let:
Length = \(2x^2 – 5xy + 3z^2\)
Breadth = \(4xy – x^2 – z^2\)
Step 2: Add Length and Breadth: \[ (2x^2 – 5xy + 3z^2) + (4xy – x^2 – z^2) \] Group like terms: \[ = (2x^2 – x^2) + (-5xy + 4xy) + (3z^2 – z^2) \\ = x^2 – xy + 2z^2 \]Step 3: Now multiply by 2 to get the perimeter: \[ \text{Perimeter} = 2 \times (x^2 – xy + 2z^2) \\ = 2x^2 – 2xy + 4z^2 \]Answer: The perimeter of the rectangle is \(2x^2 – 2xy + 4z^2\)

Q7: What must be subtracted from \(19x^4 + 2x^3 + 30x – 37\) to get \(8x^4 + 22x^3 – 7x – 60\)?

Step 1: Let the required expression be \(P(x)\).
According to the question: \[ 19x^4 + 2x^3 + 30x – 37 – P(x) = 8x^4 + 22x^3 – 7x – 60 \]Step 2: Rearranging the equation: \[ P(x) = (19x^4 + 2x^3 + 30x – 37) – (8x^4 + 22x^3 – 7x – 60) \]Step 3: Distribute the minus sign to the second polynomial: \[ P(x) = 19x^4 + 2x^3 + 30x – 37 – 8x^4 – 22x^3 + 7x + 60 \]Step 4: Group like terms: \[ = (19x^4 – 8x^4) + (2x^3 – 22x^3) + (30x + 7x) + (-37 + 60) \\ = 11x^4 – 20x^3 + 37x + 23 \]Answer: The expression to be subtracted is \(11x^4 – 20x^3 + 37x + 23\)

Q8: How much smaller is \(15x – 18y + 19z\) than \(22x – 20y – 13z + 26\)?

Step 1: Let the first expression be \(A = 15x – 18y + 19z\)
Let the second expression be \(B = 22x – 20y – 13z + 26\)
We are asked: How much smaller is \(A\) than \(B\)?
That means, compute \(B – A\)Step 2: Write the subtraction: \[ B – A = (22x – 20y – 13z + 26) – (15x – 18y + 19z) \]Step 3: Distribute the minus sign to the second expression: \[ = 22x – 20y – 13z + 26 – 15x + 18y – 19z \]Step 4: Combine like terms: \[ = (22x – 15x) + (-20y + 18y) + (-13z – 19z) + 26 \\ = 7x – 2y – 32z + 26 \]
Answer: \(15x – 18y + 19z\) is smaller than \(22x – 20y -13z + 26\) by \(7x – 2y – 32z + 26\)

Q9: How bigger is \(5x^2y^2 – 18xy^2 – 10x^2y\) than \(-5x^2 + 6x^2y – 7xy\)?

Step 1: Let the first expression be \[ A = 5x^2y^2 – 18xy^2 – 10x^2y \] and the second expression be \[ B = -5x^2 + 6x^2y – 7xy \]We are asked: How much bigger is \(A\) than \(B\)?
That means, compute \(A – B\)Step 2: Write the subtraction: \[ A – B = (5x^2y^2 – 18xy^2 – 10x^2y) – (-5x^2 + 6x^2y – 7xy) \]Step 3: Distribute the minus sign to the second expression: \[ = 5x^2y^2 – 18xy^2 – 10x^2y + 5x^2 – 6x^2y + 7xy \]Step 4: Rearranging and combining like terms:
Group similar terms:
– \(5x^2y^2\) (no like term)
– \(-18xy^2\) (no like term)
– \(-10x^2y – 6x^2y = -16x^2y\)
– \(+5x^2\)
– \(+7xy\)
So the expression becomes: \[ = 5x^2y^2 – 18xy^2 – 16x^2y + 5x^2 + 7xy \]Answer: \(5x^2y^2 – 18xy^2 – 10x^2y\) is bigger than \(-5x^2 + 6x^2y – 7xy\) by \(5x^2y^2 – 18xy^2 – 16x^2y + 5x^2 + 7xy\)

Q10: If \(x=2\) and \(y=1\); find the value of \(\left(-4x^2y^3\right)\times\left(-5x^2y^5\right)\)

Step 1: Write the given expression
⇒ \(\left(-4x^2y^3\right)\times\left(-5x^2y^5\right)\)
Step 2: Multiply the constants
⇒ \(-4 \times -5 = 20\)
Step 3: Use law of indices for same base: \(x^m \times x^n = x^{m+n}\)
⇒ \(x^2 \times x^2 = x^{2+2} = x^4\)
⇒ \(y^3 \times y^5 = y^{3+5} = y^8\)
Step 4: So the expression becomes:
⇒ \(20x^4y^8\)
Step 5: Substitute \(x=2\) and \(y=1\)
⇒ \(20 \times (2)^4 \times (1)^8\)
Step 6: Simplify powers
⇒ \(20 \times 16 \times 1 = 320\)
Answer: 320

Q11: Evaluate:

i. \((3x – 2)(x + 5)\) for \(x = 2\)

Step 1: Substitute \(x = 2\) into the expression. \[ (3x – 2)(x + 5) = (3 \cdot 2 – 2)(2 + 5) \]Step 2: Simplify inside each bracket: \[ = (6 – 2)(7) = (4)(7) \]Step 3: Multiply the two values: \[ = 28 \]Answer: 28

ii. \((2x – 5y)(2x + 3y)\) for \(x = 2\), \(y = 3\)

Step 1: Substitute \(x = 2\), \(y = 3\) into the expression: \[ (2x – 5y)(2x + 3y) = (2 \cdot 2 – 5 \cdot 3)(2 \cdot 2 + 3 \cdot 3) \]Step 2: Simplify each part: \[ = (4 – 15)(4 + 9) = (-11)(13) \]Step 3: Multiply: \[ = -143 \]Answer: -143

iii. \(xz(x^2 + y^2)\) for \(x = 2\), \(y = 1\), \(z = 1\)

Step 1: Substitute the values into the expression: \[ xz(x^2 + y^2) = 2 \cdot 1 \cdot (2^2 + 1^2) \]Step 2: Calculate the squares: \[ = 2 \cdot 1 \cdot (4 + 1) = 2 \cdot 1 \cdot 5 \]Step 3: Multiply: \[ = 10 \]Answer: 10

Q12: Evaluate:

i. \(x(x-5)+2\) for x = 1

Step 1: Write the given expression:
 \(x(x – 5) + 2\)
Step 2: Substitute x = 1 into the expression:
 \(1(1 – 5) + 2\)
Step 3: Simplify the bracket:
 \(1 \times (-4) + 2\)
Step 4: Multiply:
 \(-4 + 2\)
Step 5: Final result:
 \(-2\)
Answer: -2

ii. \(xy^2(x – 5y) + 1\) for x = 2 and y = 1

Step 1: Write the given expression:
 \(xy^2(x – 5y) + 1\)
Step 2: Substitute x = 2, y = 1:
 \(2 \cdot (1)^2 \cdot (2 – 5 \cdot 1) + 1\)
Step 3: Simplify powers and brackets:
 \(2 \cdot 1 \cdot (2 – 5) + 1\)
 \(2 \cdot (-3) + 1\)
Step 4: Multiply:
 \(-6 + 1\)
Step 5: Final result:
 \(-5\)
Answer: -5

iii. \(2x(3x – 5) – 5(x – 2) – 18\) for x = 2

Step 1: Write the given expression:
 \(2x(3x – 5) – 5(x – 2) – 18\)
Step 2: Substitute x = 2:
 \(2 \cdot 2 (3 \cdot 2 – 5) – 5 (2 – 2) – 18\)
Step 3: Simplify inside the brackets:
 \(4(6 – 5) – 5(0) – 18\)
 \(4 \cdot 1 – 0 – 18\)
Step 4: Multiply and subtract:
 \(4 – 0 – 18 = -14\)
Answer: -14

Q13: Multiply and then verify: (-3x²y²) and (x – 2y) for x = 1 and y = 2.

i. Multiply the expressions:

Step 1: Write the given expressions:
(−3x²y²)(x − 2y)
Step 2: Apply distributive property:
= (−3x²y²) × x + (−3x²y²) × (−2y)
Step 3: Multiply each term:
= −3x³y² + 6x²y³
Answer: −3x³y² + 6x²y³

ii. Verification for x = 1 and y = 2:

Step 1: Evaluate LHS:
(−3x²y²)(x − 2y)
Substitute x = 1, y = 2
= (−3 × 1² × 2²)(1 − 2×2)
= (−3 × 1 × 4)(1 − 4)
= (−12)(−3) = 36
Step 2: Evaluate RHS:
−3x³y² + 6x²y³
Substitute x = 1, y = 2
= −3 × 1³ × 2² + 6 × 1² × 2³
= −3 × 1 × 4 + 6 × 1 × 8
= −12 + 48 = 36
Answer: LHS = RHS = 36 ✅ Verified

Q14: Multiply:

i. Multiply: \( (2x^2 – 4x + 5)(x^2 + 3x – 7) \)

Step 1: Distribute each term of the first polynomial to every term of the second polynomial.
= \( 2x^2(x^2 + 3x – 7) \) + \( (-4x)(x^2 + 3x – 7) \) + \( 5(x^2 + 3x – 7) \)
Step 2: Multiply each term:
= \( 2x^4 + 6x^3 – 14x^2 \) + \( -4x^3 – 12x^2 + 28x \) + \( 5x^2 + 15x – 35 \)
Step 3: Combine like terms:
= \( 2x^4 + (6x^3 – 4x^3) + (-14x^2 – 12x^2 + 5x^2) + (28x + 15x) – 35 \)
= \( 2x^4 + 2x^3 – 21x^2 + 43x – 35 \)
Answer: 2x⁴ + 2x³ − 21x² + 43x − 35

ii. Multiply: \( (ab – 1)(3 – 2ab) \)

Step 1: Apply distributive property (FOIL method):
= \( ab × 3 + ab × (-2ab) + (-1) × 3 + (-1) × (-2ab) \)
Step 2: Multiply each term:
= \( 3ab – 2a^2b^2 – 3 + 2ab \)
Step 3: Combine like terms:
= \( (3ab + 2ab) – 2a^2b^2 – 3 \) = \( 5ab – 2a^2b^2 – 3 \)
Answer: 5ab − 2a²b² − 3

Q15: Simplify: \(\left(5 – x\right)\left(6 – 5x\right)\left(2 – x\right)\)

\(\left(5 – x\right)\left(6 – 5x\right)\left(2 – x\right)\)
Step 1: Group and simplify the first two brackets:
\((5 – x)(6 – 5x) =\)
Using distributive property:
= \(5 \cdot 6 – 5 \cdot 5x – x \cdot 6 + x \cdot 5x\)
= \(30 – 25x – 6x + 5x^2\)
= \(5x^2 – 31x + 30\)
Step 2: Multiply the result with the third bracket:
\(\left(5x^2 – 31x + 30\right)\left(2 – x\right)\)
Now apply distributive property again:
= \(5x^2 \cdot 2 – 5x^2 \cdot x – 31x \cdot 2 + 31x \cdot x + 30 \cdot 2 – 30 \cdot x\)
= \(10x^2 – 5x^3 – 62x + 31x^2 + 60 – 30x\)
Step 3: Combine like terms:
= \(-5x^3 + (10x^2 + 31x^2) + (-62x – 30x) + 60\)
= \(-5x^3 + 41x^2 – 92x + 60\)
Answer: \(-5x^3 + 41x^2 – 92x + 60\)

Q16: The product of two numbers is \(16x^4 – 1\). If one number is \(2x – 1\), find the other.

Product = \(16x^4 – 1\), one number = \(2x – 1\)
We need to find the other number.
Step 1: Recognize the expression \(16x^4 – 1\) as a difference of squares: \[ 16x^4 – 1 = \left(4x^2\right)^2 – 1^2 = \left(4x^2 – 1\right)\left(4x^2 + 1\right) \]Step 2: Again, factor \(4x^2 – 1\) using difference of squares: \[ 4x^2 – 1 = \left(2x – 1\right)\left(2x + 1\right) \]So, \[ 16x^4 – 1 = \left(2x – 1\right)\left(2x + 1\right)\left(4x^2 + 1\right) \]Step 3: Since one of the numbers is \(2x – 1\), the other must be: \[ \left(2x + 1\right)\left(4x^2 + 1\right) \\ 2x(4x^2 + 1) + 1(4x^2 + 1) = 8x^3 + 2x + 4x^2 + 1 \]Answer: \(8x^3 + 2x + 4x^2 + 1\)

Q17: Divide \(x^6 – y^6\) by the product of \(x^2 + xy + y^2\) and \(x – y\).

Step 1: The product of (x2 + xy + y2) and (x – y) \[ = (x2 + xy + y2) × (x – y) \\ = x × (x2 + xy + y2) – y × (x2 + xy + y2) \\ = x^{1+2} + x^{1+1}y + xy^2 – x^2y – xy^{1+1} – y^{1+2} \\ = x^3 + x^2y + xy^2 – x^2y – xy^2 – y^3 \\ = x^3 + (x^2y – x^2y) + (xy^2 – xy^2) – y^3 \\ = x^3 – y^3 \]Step 2:Factor of \((x^6 – y^6)\) \[ (x^6 – y^6) = x^{3 \times 2} – y^{3 \times 2} = (x^3 – y^3)(x^3 + y^3) \]Step 3:Now, \((x^6 – y^6) ÷ ( x^3 – y^3)\) \[ (x^6 – y^6) ÷ ( x^3 – y^3) = \frac{(x^3 – y^3)(x^3 + y^3)}{(x^3 – y^3)} \\ \Rightarrow (x^3 + y^3) \]Answer: \((x^3 + y^3)\)

Q18: Simplify: \(3a\times\left[8b\div4-6\left\{a-\left(5a-\overline{3b-2a}\right)\right\}\right]\)

\(3a × [8b ÷ 4 − 6 { a − (5a − \overline{3b − 2a})}]\)
Step 1: Simplify the underlined part: \(\overline{3b − 2a} = 3b − 2a\)
Step 2: Substitute back into the expression:
= 3a × [8b ÷ 4 − 6 { a − (5a − (3b − 2a)) }]
Step 3: Simplify the innermost brackets first:
= 3a × [8b ÷ 4 − 6 { a − (5a − 3b + 2a) }]
= 3a × [8b ÷ 4 − 6 { a − (7a − 3b) }]
Step 4: Continue simplifying:
= 3a × [8b ÷ 4 − 6 { a − 7a + 3b }]
= 3a × [8b ÷ 4 − 6 { −6a + 3b }]
Step 5: Simplify 8b ÷ 4 = 2b
= 3a × [2b − 6(−6a + 3b)]
Step 6: Expand −6(−6a + 3b):
= 3a × [2b + 36a − 18b]
Step 7: Simplify inside the brackets:
= 3a × [−16b + 36a]
Step 8: Multiply 3a with each term inside the bracket:
= 3a × (−16b) + 3a × (36a)
= −48ab + 108a²
Answer: −48ab + 108a²

Q19: \(7x+4\left\{x^2\div\left(5x\div 10\right)\right\}-3\left\{2-x^3\div\left(3x^2\div x\right)\right\}\)

Step 1: Write the full expression and simplify the inner divisions
= 7x + 4{ x² ÷ (5x ÷ 10) } – 3{ 2 – x³ ÷ (3x² ÷ x) }
Step 2: Simplify the divisions inside the curly brackets
First bracket:
⇒ 5x ÷ 10 = (5x / 10) = x/2
⇒ So, x² ÷ (x/2) = x² × (2/x) = 2x
Second bracket:
⇒ 3x² ÷ x = (3x² / x) = 3x
⇒ So, x³ ÷ 3x = x³ / 3x = x² / 3
Now substitute back:
= 7x + 4 × (2x) – 3 × (2 – x² / 3)
Step 3: Multiply terms
= 7x + 8x – 3 × (2 – x²/3)
Step 4: Distribute -3 in the last bracket
= 7x + 8x – [3×2 – 3×(x²/3)]
= 15x – (6 – x²)
Step 5: Remove brackets
= 15x – 6 + x²
Step 6: Rearranging in standard form
= x² + 15x – 6
Answer: x² + 15x – 6