Algebra Basics — Advanced Exam Problems

🎯 Learning Objective

Answer identity-chain questions — powers of x + 1/x, surd inputs and three-variable symmetric sums — without ever solving for x.

💡 Concept

• The x + 1/x family: every higher power comes from squaring or cubing the given value — x itself is never needed
• Surd values like 2 + √3 are disguises: 1/x rationalises to the conjugate and x + 1/x turns clean
• Three-variable bridge: (a + b + c)² links the sum, the pairwise products and the squares
• a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca); if a + b + c = 0, cubes sum to 3abc
• RRB/SSC set these as 30-second questions for identity-users and 5-minute traps for equation-solvers

🧮 Key Formulas

x² + 1/x² = (x + 1/x)² − 2

>

x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x)

>

a² + b² + c² = (a + b + c)² − 2(ab + bc + ca)

>

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

✏️ Easy Example

Q. If x + 1/x = 6, find x² + 1/x².

1. Square the given: (x + 1/x)² = x² + 2·x·(1/x) + 1/x² — the middle term collapses to plain 2 because x × 1/x = 1
2. So 36 = x² + 1/x² + 2
3. x² + 1/x² = 36 − 2 = 34

Answer: 34

🇮🇳 Real-Life Example

Computers avoid slow root-finding the same way — identity shortcuts beat brute force, in code and in the exam hall.

📝 Exam-Level Example

Q. If x + 1/x = 5, find the value of x³ + 1/x³.

1. Use (a + b)³ = a³ + b³ + 3ab(a + b) with a = x and b = 1/x — here ab = x × 1/x = 1, so the 3ab term becomes just 3
2. So (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x)
3. Substitute the given value: 5³ = x³ + 1/x³ + 3 × 5
4. 125 = x³ + 1/x³ + 15
5. x³ + 1/x³ = 125 − 15 = 110

Answer: 110

📝 Exam-Level Example

Q. If x = 2 + √3, find the value of x² + 1/x².

1. First find 1/x: 1/(2 + √3) — multiply top and bottom by (2 − √3) to rationalise the denominator
2. 1/x = (2 − √3)/((2)² − (√3)²) = (2 − √3)/(4 − 3) = 2 − √3
3. So x + 1/x = (2 + √3) + (2 − √3) = 4 — the surds cancel, which is the whole design of the question
4. Now apply the ladder: x² + 1/x² = (x + 1/x)² − 2 = 4² − 2
5. = 16 − 2 = 14

Answer: 14

📝 Exam-Level Example

Q. If a + b + c = 9 and ab + bc + ca = 26, find a² + b² + c² and the value of a³ + b³ + c³ − 3abc.

1. Square the sum: (a + b + c)² = a² + b² + c² + 2(ab + bc + ca) — the bridge between the two given facts
2. 81 = (a² + b² + c²) + 2 × 26 = (a² + b² + c²) + 52
3. a² + b² + c² = 81 − 52 = 29
4. Factor identity: a³ + b³ + c³ − 3abc = (a + b + c)(a² + b² + c² − ab − bc − ca)
5. = 9 × (29 − 26) = 9 × 3 = 27

Answer: a² + b² + c² = 29; a³ + b³ + c³ − 3abc = 27

🪄 Memory Trick

The x + 1/x ladder: k → k² − 2 → k³ − 3k. Climb step by step; never solve for x itself.

⚠️ Common Mistakes

• ❌ Solving the quadratic for x — identities finish in two lines, roots waste five minutes
• ❌ Carrying ab as an unknown — for x and 1/x, the product is always exactly 1
• ❌ Sign slip with x − 1/x = k: squaring gives x² + 1/x² = k² + 2 (plus, not minus)

🏆 Exam Tips

• ✅ If a + b + c = 0 appears anywhere, then a³ + b³ + c³ = 3abc instantly
• ✅ Surd inputs like 2 + √3 almost always rationalise to a clean x + 1/x
• ✅ Memorise ladder values: k = 3 → 7, 18; k = 4 → 14, 52; k = 5 → 23, 110

📌 Summary

• Never find x — climb the identity ladder
• Rationalise surds to expose x + 1/x
• (a + b + c)² bridges sum, products and squares
• a + b + c = 0 → cubes sum to 3abc