1. Reciprocal Powers: x + 1/x

What is the question?

These questions usually give:

x + 1/x = some value

and ask you to find:

x² + 1/x²
x³ + 1/x³
x⁴ + 1/x⁴

You usually do not need to find x.

Idea

Start with:

x + 1/x

Square it:

(x + 1/x)²
= x² + 2 + 1/x²

Therefore:

x² + 1/x²
= (x + 1/x)² - 2

For cubes:

x³ + 1/x³
= (x + 1/x)³ - 3(x + 1/x)

Example

Question: If:

x + 1/x = 4

find:

x³ + 1/x³

Step 1: Use the formula

x³ + 1/x³
= (x + 1/x)³ - 3(x + 1/x)

Step 2: Substitute 4

= 4³ - 3(4)

Step 3: Calculate

= 64 - 12
= 52

Answer: 52

Remember

x + 1/x = a

x² + 1/x² = a² - 2

x³ + 1/x³ = a³ - 3a

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2. Symmetric Cubic Identity

What is the question?

These questions usually contain three variables:

a, b, c

and give a condition such as:

a + b + c = 0

They then ask about:

a³ + b³ + c³

or:

abc

Idea

The important identity is:

a³ + b³ + c³ - 3abc
=
(a+b+c)(a²+b²+c²-ab-bc-ca)

The important special case is:

a + b + c = 0

Then:

a³ + b³ + c³ = 3abc

Example

Question: If:

a + b + c = 0

find:

(a³ + b³ + c³) / abc

Step 1: Use the special condition

Since:

a + b + c = 0

we know:

a³ + b³ + c³ = 3abc

Step 2: Substitute

(a³ + b³ + c³) / abc

= 3abc / abc

= 3

Answer: 3

Shortcut

Whenever you see:

a + b + c = 0

immediately think:

a³ + b³ + c³ = 3abc

This is the main pattern you need for aptitude exams.

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3. Polynomial Root Relations

What is the question?

These questions give a quadratic equation and say that its roots are:

α and β

Then they ask you to find something like:

α + β
αβ
α² + β²
α³ + β³
1/α + 1/β

Idea

You usually do not need to solve the quadratic.

For:

ax² + bx + c = 0

if the roots are α and β:

α + β = -b/a

αβ = c/a

These two values can be used to find many other expressions.

Example

Question: If α and β are roots of:

x² - 5x + 6 = 0

find:

α² + β²

Step 1: Find the sum

Here:

a = 1
b = -5
c = 6

Therefore:

α + β = -(-5)/1
      = 5

Step 2: Find the product

αβ = 6/1
   = 6

Step 3: Use the identity

α² + β²
= (α + β)² - 2αβ

Step 4: Substitute

= 5² - 2(6)

= 25 - 12

= 13

Answer: 13

Remember

For:

ax² + bx + c = 0

always remember:

Sum of roots    = -b/a
Product of roots = c/a

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4. Maximum and Minimum Using AM-GM

What is the question?

These questions ask:

“Find the minimum value of an expression.”

or:

“Find the maximum/minimum possible value.”

A common aptitude pattern is:

x + 1/x

or:

ax + b/x

where x > 0.

Idea

For positive numbers:

(a+b)/2 ≥ √ab

Therefore:

a+b ≥ 2√ab

This is called AM-GM.

Example

Question: Find the minimum value of:

x + 1/x

where:

x > 0

Step 1: Apply AM-GM

x + 1/x ≥ 2√(x × 1/x)

Step 2: Simplify

= 2√1
= 2

Therefore:

x + 1/x ≥ 2

Answer: Minimum = 2

When does the minimum occur?

AM-GM gives equality when the two terms are equal:

x = 1/x

Therefore:

x² = 1

Since:

x > 0

we get:

x = 1

Shortcut

For:

x + 1/x, x > 0

remember:

Minimum = 2

More generally:

ax + b/x, x > 0

has minimum:

2√ab

when a,b > 0.

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5. Factor Theorem and Remainder Theorem

What is the question?

These questions usually ask:

“Find the remainder when a polynomial is divided by x-a.”

Or:

“Check whether x-a is a factor of the polynomial.”

Idea

You don’t need to perform polynomial long division.

If:

P(x)

is divided by:

x-a

the remainder is:

P(a)

This is the Remainder Theorem.

If:

P(a) = 0

then:

x-a

is a factor.

This is the Factor Theorem.

Example

Question: Find the remainder when:

P(x) = x³ - 2x² + x - 5

is divided by:

x - 2

Step 1: Identify a

x - 2 = x - a

Therefore:

a = 2

Step 2: Find P(2)

P(2)
= 2³ - 2(2²) + 2 - 5

Step 3: Calculate

= 8 - 8 + 2 - 5

= -3

Answer: Remainder = -3

Factor example

Question: Is x-2 a factor of:

P(x) = x³ - 5x² + 8x - 4

Check:

P(2)
= 8 - 20 + 16 - 4
= 0

Therefore:

x - 2 is a factor.

Shortcut

Divided by x-a
       ↓
Find P(a)

P(a) = 0
       ↓
x-a is a factor

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6. Algebraic Inequalities and Modulus

What is the question?

These questions usually ask:

“Find the range of x.”

or:

“Solve the inequality.”

When modulus appears, you’ll see:

|x|

Idea

Absolute value means distance from zero.

For example:

|5| = 5
|-5| = 5

Therefore:

|x| ≤ a

means:

x is at most a units away from zero.

So:

-a ≤ x ≤ a

Similarly:

|x| ≥ a

means:

x ≤ -a OR x ≥ a

Example

Question: Solve:

|2x - 3| ≤ 5

Step 1: Remove the modulus

Use:

|A| ≤ k
→ -k ≤ A ≤ k

Therefore:

-5 ≤ 2x - 3 ≤ 5

Step 2: Add 3 everywhere

-2 ≤ 2x ≤ 8

Step 3: Divide by 2

-1 ≤ x ≤ 4

Answer:

-1 ≤ x ≤ 4

Shortcut

Remember:

|A| ≤ k
→ -k ≤ A ≤ k

and:

|A| ≥ k
→ A ≤ -k OR A ≥ k

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7. Higher Reciprocal Powers

What is the question?

These questions give:

x + 1/x

and ask for a higher power such as:

x⁴ + 1/x⁴
x⁶ + 1/x⁶