NDA Algebraic Identities — Study Material, 31 PYQs & Practice MCQs | ZestExam
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NDA Algebraic Identities
Study Material — 31 PYQs (2018–2021) · Concept Notes · Shortcuts
NDA Algebraic Identities is a frequently tested subtopic — 31 previous year questions from 2018–2021 papers are included below with concept notes, key rules and shortcut tricks.
If x + y = 12 and xy = 35, find the value of x² + y².
Exam Q 42021Previous Year Pattern
If a - b = 5 and ab = 24, find the value of a² + b².
Exam Q 52021Previous Year Pattern
Expand: (2x + 3y)(2x - 3y)
Exam Q 62021Previous Year Pattern
Simplify: (m + n)³ - (m - n)³
Exam Q 72021Previous Year Pattern
If a - b = 5 and ab = 24, then find the value of a³ - b³.
Exam Q 82021Previous Year Pattern
Simplify: (a + b)³ - (a - b)³
Exam Q 92021Previous Year Pattern
If p + q + r = 15 and pq + qr + rp = 72, then find the value of p² + q² + r².
Exam Q 102021Previous Year Pattern
If x + y = 12 and xy = 35, then find the value of x² + y².
Exam Q 112018Previous Year Pattern
If x + y = 10 and xy = 21, find the value of x² + y².
Exam Q 122021Previous Year Pattern
If p + q + r = 15 and pq + qr + rp = 72, find the value of p² + q² + r².
Exam Q 132021Previous Year Pattern
Simplify: (a + b)³ - (a - b)³.
Exam Q 142018Previous Year Pattern
If x + 1/x = 5, then find the value of x² + 1/x².
Exam Q 152021Previous Year Pattern
If a² + b² = 13 and ab = 6, find the value of (a + b)².
Exam Q 162021Previous Year Pattern
If x - y = 4 and xy = 5, find the value of x³ - y³.
Exam Q 172021Previous Year Pattern
Simplify: (2a + 3b)² - (2a - 3b)².
Exam Q 182021Previous Year Pattern
If x² + y² = 34 and xy = 15, find the value of (x + y)².
Exam Q 192021Previous Year Pattern
If a - b = 3 and ab = 4, find the value of a³ - b³.
Exam Q 202021Previous Year Pattern
If a + b + c = 9 and ab + bc + ca = 26, find the value of a² + b² + c².
Exam Q 212020Previous Year Pattern
If x + 1/x = 5, then find the value of x³ + 1/x³.
Exam Q 222021Previous Year Pattern
If (x + y)³ = 216 and (x − y)³ = 64, then find the value of xy.
Exam Q 232021Previous Year Pattern
If (a + b + c)² = 81 and a² + b² + c² = 35, then find the value of (a + b − c)² + (b + c − a)² + (c + a − b)².
Exam Q 242021Previous Year Pattern
If a + b + c = 12 and a² + b² + c² = 56, then find the value of ab + bc + ca.
Exam Q 252021Previous Year Pattern
If (a + b + c)² = 81 and a² + b² + c² = 35, then find the value of (a + b + c)² − 3(ab + bc + ca).
Exam Q 262021Previous Year Pattern
If x² + y² = 13 and xy = 6, then find the value of (x + y)² − (x − y)².
Exam Q 272021Previous Year Pattern
If a³ + b³ = 728 and a + b = 8, then find the value of ab.
Exam Q 282021Previous Year Pattern
If x + 1/x = 5, then find the value of x⁴ + 1/x⁴.
Exam Q 292020Previous Year Pattern
If (a + b + c)² = 81 and a² + b² + c² = 29, then find the value of (ab + bc + ca).
Exam Q 302018Previous Year Pattern
If x + 1/x = 5, then find the value of x⁵ + 1/x⁵.
Exam Q 312021Previous Year Pattern
If a + b + c = 12 and a² + b² + c² = 50, then find the value of ab + bc + ca.
Concept Notes
Algebraic Identities— Rules & Concept
💡
Core Concept
Read this first — the foundation of the topic
💡Key Rules and Properties
Algebraic identities are universally true for all values of variables. They help expand, factorize, and simplify expressions. The most important ones follow specific patterns that SSC loves to test
🔑Essential Formulas
• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²
• (a + b)(a - b) = a² - b²
• (a + b)³ = a³ + 3a²b + 3ab² + b³
• (a - b)³ = a³ - 3a²b + 3ab² - b³
• a³ + b³ = (a + b)(a² - ab + b²)
• a³ - b³ = (a - b)(a² + ab + b²)
• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
📊
Exam Patterns
What examiners ask — read before attempting PYQs
SSC CGL typically asks direct substitution problems, value finding questions, and simplification using identities. Common question types include finding the value of expressions like x² + 1/x² when x + 1/x is given, or simplifying complex algebraic fractions.
Shortcut Trick - The 'Recognition Method':
Instead of expanding everything, learn to recognize patterns. If you see (something)² - (something else)², immediately think a² - b² = (a+b)(a-b). This saves 2-3 minutes per question.
✏️
Worked Example
Solve this step-by-step before moving on
1
Step 1
Start with the given condition x + 1/x = 5
2
Step 2
Square both sides: (x + 1/x)² = 5² = 25
3
Step 3
Apply the identity (a + b)² = a² + 2ab + b²
Here, a = x and b = 1/x
4
Step 4
(x + 1/x)² = x² + 2(x)(1/x) + (1/x)²
5
Step 5
25 = x² + 2(1) + 1/x²
6
Step 6
25 = x² + 2 + 1/x²
7
Step 7
x² + 1/x² = 25 - 2 = 23
Therefore, x² + 1/x² = 23.
Common Mistake:
Students often forget the middle term in expansions. In (a + b)², many write a² + b² and miss the 2ab term. Always count terms: square of binomial has THREE terms, cube has FOUR terms. Also, students confuse signs in (a - b) expansions. Remember: alternate signs in (a - b)³.
Key Points to Remember
(a + b)² = a² + 2ab + b² - never forget the middle term 2ab
(a - b)² = a² - 2ab + b² - note the minus sign before 2ab
(a + b)(a - b) = a² - b² - difference of squares identity
a³ + b³ = (a + b)(a² - ab + b²) - sum of cubes factorization
a³ - b³ = (a - b)(a² + ab + b²) - difference of cubes factorization
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca - three variable expansion
When x + 1/x is given, find x² + 1/x² by squaring both sides
Recognize patterns first, then apply appropriate identity to save time
Exam-Specific Tips
(a + b)² has exactly 3 terms: a², 2ab, and b²
(a - b)³ = a³ - 3a²b + 3ab² - b³ with alternating signs