Algebra Exercises

Cubes, Expansion, Factorization

1️. Cubes Using Formula

Use the identity:

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

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

Examples:

1. $(x + 2)^3$

   → $x^3 + 3x^2 \cdot 2 + 3x \cdot 4 + 8$

   = $x^3 + 6x^2 + 12x + 8$

2. $(3a - 5)^3$

   → $27a^3 - 135a^2 + 225a - 125$

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Write the Expressions in Expanded Form

Examples:

1. Expand $(2x - 1)^3$

   → $8x^3 - 12x^2 + 6x - 1$

2. Expand $(a + \dfrac{1}{2})^3$

   → $a^3 + \dfrac{3}{2}a^2 + \dfrac{3}{4}a + \dfrac{1}{8}$

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Write in the Form of $(a + b)^3$ or $(a - b)^3$

Examples:

1. $x^3 + 6x^2 + 12x + 8$

   → $(x)^3 + 3 \cdot (x)^2 \cdot (2) + 3 \cdot (x) \cdot (2)^2 + (2)^3$

   → $(x + 2)^3$

2. $27a^3 - 135a^2 + 225a - 125$

   → $(3a)^3 - 3 \cdot (3a)^2 \cdot (5) + 3 \cdot (3a) \cdot (5) ^2 - (5)^3$

   → $(3a - 5)^3$

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Factorize (Including Fractional Coefficients)

Examples:

1. Factorize: $x^3 + \dfrac{3}{2}x^2 + \dfrac{3}{4}x + \dfrac{1}{8}$

   → Recognize as $(x + \dfrac{1}{2})^3$

2. Factorize: $8a^3 - 27b^3$

   → Use difference of cubes: $(2a)^3 - (3b)^3$

   = $(2a - 3b)(4a^2 + 6ab + 9b^2)$

3. Factorize: $\dfrac{1}{8}x^3 + \dfrac{3}{4}x^2 + \dfrac{3}{2}x + 1$

   → Recognize as $\left( \dfrac{1}{2}x + 1 \right)^3$

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Factorize the expression, $a^4 + a^2b^2 + b^4$.

Here, = $(a^2)^2 + 2a^2b^2 – a^2b^2 + (b^2)^2$

= $(a^2 + b^2)^2 – (ab)^2 [⸪ (a + b)^2 = a^2 + 2ab + b^2]$

= $(a^2 + b^2 + ab) (a^2 + b^2 – ab) [⸪ a^2 – b^2 = (a + b) (a – b)]$

= $(a^2 + ab + b^2) (a^2 – ab + b^2)$

$\therefore a^4 + a^2b^2 + b^4 = (a^2 + ab + b^2) (a^2 – ab + b^2)$

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Factorization Examples

Factorize: $y^4 + y^2 + 1$

Treat as quadratic in $y^2$.

$y^4 + y^2 + 1$

= $y^4 + 2y^2 - y^2 + 1$

= $y^4 + 2y^2 + 1 - y^2$

= $(y^2 + 1)^2 - y^2$

= $(y^2 + y + 1)(y^2 - y + 1)$

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Factorize: $49a^4 - 154a^2b^2 + 9b^4$

Observe a "squared minus square" form:

$49a^4 - 154a^2b^2 + 9b^4$

= $(7a^2 + 3b^2)^2 - (14ab)^2$

= $(7a^2 + 14ab + 3b^2)(7a^2 - 14ab + 3b^2)$

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Factorize: $\dfrac{x^4}{y^4} + \dfrac{x^2}{y^2} + 1$

Let $z=\dfrac{x}{y}$.

Then, $\dfrac{x^4}{y^4} + \dfrac{x^2}{y^2} + 1$

= $z^4 + z^2 + 1 = (z^2 + z + 1)(z^2 - z + 1)$.

Substituting $z=\dfrac{x}{y}$

= $(\dfrac{x^2}{y^2} + \dfrac{x}{y} + 1) (\dfrac{x^2}{y^2} - \dfrac{x}{y} + 1)$

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HCF and LCM – Examples

Example: Find the HCF of 24 and 36

Method: Prime Factorization

• $24 = 2^3 \cdot 3$

• $36 = 2^2 \cdot 3^2$

HCF = Product of lowest powers of common primes:

$= 2^2 \cdot 3 = 12$

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Example: Find the LCM of 24 and 36

Method: Prime Factorization

• $24 = 2^3 \cdot 3$

• $36 = 2^2 \cdot 3^2$

LCM = Product of highest powers of all primes:

$= 2^3 \cdot 3^2 = 72$

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Example: Find HCF and LCM of 18 and 30

• $18 = 2 \cdot 3^2$

• $30 = 2 \cdot 3 \cdot 5$

HCF = $2 \cdot 3 = 6$

LCM = $2 \cdot 3^2 \cdot 5 = 90$

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Example: Relationship Between HCF and LCM

Question:

Find the HCF and LCM of 12 and 15. Verify:

$\text{HCF} \cdot \text{LCM} = \text{Product of numbers}$

• $12 = 2^2 \cdot 3$

• $15 = 3 \cdot 5$

HCF = $3$

LCM = $2^2 \cdot 3 \cdot 5 = 60$

Check:
$3 \cdot 60 = 180$

$12 \cdot 15 = 180$

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Example: Word Problem Using HCF

Question:

Two ropes are 60 cm and 84 cm long. What is the greatest length that can be used to cut both ropes without remainder?

Solution:
Find HCF of 60 and 84:

• $60 = 2^2 \cdot 3 \cdot 5$

• $84 = 2^2 \cdot 3 \cdot 7$

HCF = $2^2 \cdot 3 = 12$ cm

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Example: Word Problem Using LCM

Question:

Two traffic lights blink every 45 seconds and 60 seconds respectively. After how many seconds will they blink together?

Solution:
Find LCM of 45 and 60:

• $45 = 3^2 \cdot 5$

• $60 = 2^2 \cdot 3 \cdot 5$

LCM = $2^2 \cdot 3^2 \cdot 5 = 180$ seconds

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Example: Fractional HCF and LCM

Question:

Find HCF and LCM of $\frac{3}{4}$ and $\frac{5}{6}$

Method:

• HCF = $\dfrac{\text{HCF of numerators}}{\text{LCM of denominators}}$

• LCM = $\dfrac{\text{LCM of numerators}}{\text{HCF of denominators}}$

Solution:

• HCF of 3 and 5 = 1

• LCM of 4 and 6 = 12

  → HCF = $\dfrac{1}{12}$

• LCM of 3 and 5 = 15

• HCF of 4 and 6 = 2

  → LCM = $\dfrac{15}{2}$

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Linear Equations

Word Problems – Simultaneous Equations

Example 1: Age Problem

Question: A father is 30 years older than his son. Their combined age is 50. Find their ages.

Let:

$x$ = son's age

$x + 30$ = father's age

Equation:
$x + (x + 30)$

= $50 \Rightarrow 2x + 30$

= $50 \Rightarrow x$

= $10$

Father's age = $10 + 30 = 40$

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Example 2: Money Problem

Question: A man has Rs. 100 in 5 and 10 rupee notes. He has 12 notes in total. How many of each?

Let:

• $x$ = number of Rs. 5 notes

• $y$ = number of Rs. 10 notes

Equations:

• $x + y = 12$

• $5x + 10y = 100$

Solve using substitution or elimination:

• Multiply first by 5: $5x + 5y = 60$

• Subtract: $(5x + 10y) - (5x + 5y)$

= $100 - 60 \Rightarrow 5y$

= $40$

$\Rightarrow y = 8$

$\therefore x = 12 - 8 = 4$

Answer: There are 4 notes of Rs. 5 and 8 notes of Rs. 10

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Example 3: Geometry Problem

Question: The perimeter of a rectangle is 40 cm. Its length is 4 cm more than its breadth. Find dimensions.

Let:

• $x$ = breadth
• $x + 4$ = length

Perimeter: $2(x + x + 4)$

= $40 \Rightarrow 2(2x + 4)$

= $40 \Rightarrow 4x + 8$

= $4x = 40 -8 = 32$

= $x = \dfrac{32}{4}$

$\Rightarrow x = 8$

Length = $8 + 4 = 12$

Answer: Length = 12 cm, Breadth = 8 cm