Everything the Algebraic Result chapter is asked on in SSC CGL, CHSL, CPO and other competitive exams — the ladder that takes x + 1/x up to any power, the cube and sixth-power rules, powers the ladder cannot reach, the five disguises that hide the reciprocal pair, and the sign and range rules that decide the final answer. Every concept used in the chapter test is explained here from zero.
The x + 1/x family is one of the most predictable sources of marks in SSC quantitative aptitude. Almost every shift carries a question from it — a reciprocal sum that must become a cube, a quadratic that hides the pair, a surd value that turns whole the moment you cross over, or a power so large that only a cycle can reach it. None of them needs heavy calculation. Each is a walk up or down one short ladder.
These notes explain every rule from the beginning — not just what it is but why it works, because the “why” is what lets you handle a disguise you have not seen before. Every rule comes with a worked example, an exam shortcut, and the mistake that costs the most marks.
Notes fix the method; only practice fixes the speed. Keep these open in one tab and a test in another — start with this chapter's own test, then move to full papers once the rules feel automatic.
Two levels matched to these notes. Every question has a step-by-step solution, in Hindi too.
Start practising Chapter testThe companion chapter. The identities behind the ladder are set out there in full.
Start practising Previous yearReal shift-wise CGL papers with answer keys and full solutions. The best guide to what is actually asked.
Solve PYQs Previous yearShift-wise CHSL papers with solutions. Useful for CHSL, CPO and MTS candidates alike.
Solve PYQs Full lengthFull-length papers in real exam timing, with current affairs and a detailed solution after every attempt.
Take a mock testHow to use these notes: read one section, cover the example with your hand, solve it yourself, then check. A rule you have only read is gone by evening. A rule you have used once stays.
This whole chapter is one family of questions. You are given something about x and its reciprocal 1/x, and asked for a higher power of the same pair. A small ladder of rules takes you up or down one level at a time.
x × (1/x) = 1
Because that product is exactly 1, every middle term in every expansion becomes a plain number — 2 or 3 — instead of an unknown. That is the entire reason this chapter works.
x2 + 1/x2 = (x + 1/x)2 − 2
x2 + 1/x2 = (x − 1/x)2 + 2
Square the bracket you were given, then adjust by 2.
Sum → subtract 2. Difference → add 2. That single sentence covers a large share of the chapter, and the same pattern returns for the cubes.
The identity does not care what sits inside. Replace x by x2 and it still works, so each squaring doubles the power:
x + 1/x → x2 + 1/x2 → x4 + 1/x4 → x8 + 1/x8, subtracting 2 each time.
If x + 1/x = 3, find x8 + 1/x8.
9 − 2 = 7, then 49 − 2 = 47, then 2209 − 2 = 2207.
The −2 applies at every step, not just the first. Most wrong options in this chapter are simply the value one step short of the answer.
Run it backwards: add 2 and take the square root.
Example. If x4 + 1/x4 = 47 and x > 0, then 47 + 2 = 49 gives x2 + 1/x2 = 7, and 7 + 2 = 9 gives x + 1/x = 3.
Add 2 if you want a sum; subtract 2 if you want a difference. The level above is the same either way — only the final step differs.
√x and 1/√x also multiply to 1, so they are an ordinary pair. Squaring them lands on x and 1/x.
Example. If √x + 1/√x = 4, then x + 1/x = 16 − 2 = 14.
x3 + 1/x3 = k3 − 3k, where k = x + 1/x
x3 − 1/x3 = k3 + 3k, where k = x − 1/x
It is the ordinary cube identity (a+b)3 = a3 + b3 + 3ab(a+b) with ab = 1. The correction term 3ab(a+b) collapses to 3k, which is why the rule is so short.
Sum subtracts 3k; difference adds 3k. Exactly the same pattern as the squares — sums subtract, differences add.
If x + 1/x = 4: 64 − 12 = 52.
If x − 1/x = 4: 64 + 12 = 76.
The sixth power is the square of the cube, so it takes two moves, not five.
Example. If x + 1/x = 4, the cube is 52, and x6 + 1/x6 = 522 − 2 = 2702.
If the cube you squared was a difference, the adjustment flips to +2, because you are squaring a difference bracket.
Going through the second power instead of the cube takes three steps and usually produces x4 + 1/x4 by mistake — which is always one of the options.
Odd powers above 3 are not on the doubling ladder. Build them by multiplying two levels and removing what is left over.
(x2+1/x2)(x3+1/x3) = x5+1/x5 + x + 1/x
(x3+1/x3)(x4+1/x4) = x7+1/x7 + x + 1/x
Multiplying levels m and n produces level m+n and level m−n. So 2×3 gives 5 and 1; 3×4 gives 7 and 1. Subtract the smaller level to be left with the one you want.
If x + 1/x = 3, find x5 + 1/x5.
Level 2 = 7, level 3 = 18. Then 7 × 18 − 3 = 126 − 3 = 123.
The plain product is always an option. Here 126 is offered. The leftover level must be subtracted.
Most harder questions do not hand you x + 1/x directly. They hide it. These are the five disguises SSC uses.
If the equation has 1 as its constant term, divide every term by x.
x2 − 9x + 1 = 0 → x − 9 + 1/x = 0 → x + 1/x = 9.
Constant +1 gives a sum; constant −1 gives a difference. Check the constant before you write anything down.
If the equation carries a common factor (5x2 + 5 = 40x), divide the whole equation by it first so the constant becomes 1. If it is written backwards (x2 = 8x − 1), rearrange first.
Split it. (x2+1)/x is just x + 1/x written on one line, and (x4+1)/x2 is x2 + 1/x2.
Example. (x2 + 3x + 1)/x = 7 splits to x + 3 + 1/x = 7, so x + 1/x = 4.
Rationalise. When the conjugate product is 1, the reciprocal is the conjugate, and the roots cancel on adding.
Example. x = 5 + 2√6. Since 25 − 24 = 1, 1/x = 5 − 2√6, so x + 1/x = 10.
Any pair whose product is 1 behaves exactly like x and 1/x. Name it u and carry on.
For a shifted bracket, move the constant across first. From x + 1/(x − 3) = 10, subtract 3 to get (x − 3) + 1/(x − 3) = 7. Using 10 directly is the standard error, and the wrong answer it produces is always an option.
x = 1/(x − 4) → x2 − 4x − 1 = 0 → divide by x → x − 1/x = 4.
The constant term also sets the numerator: from x = 2/(x − 3) you get x − 2/x = 3.
(x − 1/x)2 = (x + 1/x)2 − 4
The gap between the two squared brackets is always 4. This is how a surd given can turn into a whole number.
Example. x + 1/x = √29 gives (x − 1/x)2 = 29 − 4 = 25, so the difference is 5.
A square root has two signs. The question always tells you which to take.
| Condition | Then x − 1/x is |
|---|---|
| x > 1 | positive (x is the bigger term) |
| 0 < x < 1 | negative (1/x is bigger) |
| x < −1 | negative |
A sum of a square and its reciprocal, like x2 + 1/x2, is always positive whatever x is. So when you come down a ladder, the intermediate root is always the positive one — only the final step to a difference looks at the range. Mixing these two up is the commonest error in the harder questions.
For real x, x + 1/x is at least 2 or at most −2. So a given of 1 or −1 has no real solution — but the identity still fixes the answer, and SSC asks these on purpose.
Example. x + 1/x = 1 gives x3 + 1/x3 = 1 − 3 = −2.
At the other extreme, k = 2 forces x = 1, so every power equals 2. It is a useful sanity check.
When the power is huge (x50, x99), no ladder will reach it. Look for a repeat instead.
x2 + x + 1 = 0 ⇒ x3 = 1 — powers repeat every 3
x2 − x + 1 = 0 ⇒ x3 = −1, so x6 = 1 — repeat every 6
If x2 + x + 1 = 0, find x99 + 1/x99.
99 is a multiple of 3, so x99 = 1 and the answer is 2.
For x100 the remainder is 1, so it reduces to x + 1/x, which is −1.
If 9(x2 + 1/x2) = 306 and x > 0, find x + 1/x.
Strip the 9 first: x2 + 1/x2 = 34. Then 34 + 2 = 36, so the answer is 6.
If x2 + 1 = 9x, find x4 + 1/x4.
Rearrange and divide: x + 1/x = 9. Then 81 − 2 = 79, and 792 − 2 = 6239.
If x2 + 1/x2 = 27 and 0 < x < 1, find x − 1/x.
27 − 2 = 25, so the size is 5. But x is small and 1/x is large, so the answer is −5.
If x + 1/x = √13 and x > 1, find x3 − 1/x3.
Cross over: 13 − 4 = 9, so x − 1/x = 3. Then 27 + 9 = 36.
The surd disappears because 13 was chosen so that 13 − 4 is a perfect square.
If x − 1/x = 3 and x > 0, find x4 − 1/x4.
x2 + 1/x2 = 11, and (x + 1/x)2 = 9 + 4 = 13, so x + 1/x = √13.
x2 − 1/x2 = 3√13, so the answer is 3√13 × 11 = 33√13.
Any difference of even powers factorises: x4 − 1/x4 = (x − 1/x)(x + 1/x)(x2 + 1/x2). Build each bracket and multiply.
If x + 5/x = 6, find (x2 + 9x + 5) ÷ (x2 + 4x + 5).
Divide top and bottom by x: (x + 9 + 5/x) ÷ (x + 4 + 5/x) = (6 + 9) ÷ (6 + 4) = 15/10 = 3/2.
The constant 5 in each quadratic is what makes 5/x appear, matching the given exactly.
| Rule | Given → wanted | Watch out for |
|---|---|---|
| x2+1/x2 = k2 − 2 | sum → next level | −2 at every step, not just the first |
| x2+1/x2 = k2 + 2 | difference → next level | difference adds; sum subtracts |
| add 2, take root | coming back down | subtract 2 instead if you want a difference |
| x3+1/x3 = k3 − 3k | sum → cube | the 3k is easy to drop |
| x3−1/x3 = k3 + 3k | difference → cube | correction is added here |
| square the cube | → sixth power | two moves, not five |
| level m × level n | → levels m+n and m−n | subtract the leftover level |
| (x−1/x)2 = (x+1/x)2 − 4 | crossing sum ↔ difference | can turn a surd into a whole number |
| divide the quadratic by x | equation → ladder | constant must be 1; clear factors first |
| x3 = 1 or −1 | huge powers | reduce the exponent by the cycle |
Every question in this chapter is a walk up or down one ladder, wearing one of five disguises. Learn the ladder, learn the disguises, and the chapter is finished.
↑ Back to topSquare the given value and subtract 2. If the given is a difference instead, square it and add 2. The adjustment is always exactly 2 because x × 1/x = 1.
k3 − 3k, where k = x + 1/x. For the difference it is k3 + 3k with k = x − 1/x. Sums subtract the correction, differences add it.
They are not on the doubling ladder. Multiply two levels and subtract the leftover: level 2 times level 3 gives level 5 plus level 1, and level 3 times level 4 gives level 7 plus level 1.
When 0 < x < 1, because then 1/x is the larger term. It is also negative for x < −1. A sum such as x2 + 1/x2 is always positive, so only the final step to a difference depends on the range.
Look for a cycle. x2 + x + 1 = 0 means x3 = 1, so powers repeat every 3. x2 − x + 1 = 0 means x3 = −1 and x6 = 1, so they repeat every 6. Reduce the exponent by the cycle and evaluate whatever level is left.
Take the matching Algebraic Result practice test on TrickySSC after finishing a section here. The tests come in two levels and every question carries a step-by-step solution in both English and Hindi.
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