Everything the Algebraic Formula chapter is asked on in SSC CGL, CHSL, CPO and other competitive exams — the two basic squares and the four bridges they create, difference of squares, cube identities and their sign rule, three-variable identities including the a + b + c = 0 shortcut, completing the square, and factorisation. Every concept used in the chapter test is explained here from zero.
Algebraic identities are among the most dependable marks in SSC quantitative aptitude. Almost every shift carries one or two questions from this chapter — a sum and product that must become a cube, three numbers that quietly add to zero, a decimal fraction that collapses the moment you spot the right identity. None of them needs heavy calculation. Each one needs a single rule applied in the right order, in well under a minute.
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 an identity you have never seen in that exact disguise. 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 cube rule you just learned carries straight over to x + 1/x powers.
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.
Every identity in this chapter grows out of two lines. Learn them so well that you can write either one without pausing, because the rest of the chapter is these two rearranged.
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
Only the middle sign changes. The two end terms stay positive in both, because a square is never negative.
An identity is a statement that is true for every value of the letters. That is why you are almost never asked to find x in this chapter — you are asked to build one expression out of another without ever knowing what the letters are.
Rearranging those two lines gives the four results that carry most of the chapter.
| You are given | You want | Use |
|---|---|---|
| a + b and ab | a2 + b2 | (a + b)2 − 2ab |
| a − b and ab | a2 + b2 | (a − b)2 + 2ab |
| a + b and a2 + b2 | ab | [(a + b)2 − (a2 + b2)] ÷ 2 |
| a − b and a2 + b2 | ab | [(a2 + b2) − (a − b)2] ÷ 2 |
If a + b = 8 and ab = 15, find a2 + b2.
= 82 − 2(15) = 64 − 30 = 34.
Check: sum 8 and product 15 mean the numbers are 5 and 3, and 25 + 9 = 34.
(a + b)2 + (a − b)2 = 2(a2 + b2)
(a + b)2 − (a − b)2 = 4ab
Add them and the middle terms cancel, so the product never enters the working. Subtract them and the squares cancel, leaving only the product.
Add → the product disappears. Subtract → only the product survives. If a question gives you both squared brackets, you never need to find a or b separately.
Coefficients multiply into the constant. For (3a + 5b) and (3a − 5b) the difference of the squares is 4 × 3a × 5b = 60ab, not 4ab. If the question then asks for the plain product ab, divide the coefficients back out.
a2 − b2 = (a + b)(a − b)
This is the most useful single line in the chapter, because it turns a subtraction into a multiplication — and multiplications cancel against whatever else is in the question.
Whenever you see two squares with a minus between them, do not compute either square. Factorise first and see what cancels.
Find [(213)2 − (187)2] ÷ 26.
Sum = 400, difference = 26, so the numerator is 400 × 26.
The 26 cancels and the answer is 400. The divisor was placed there precisely to cancel one factor.
Each use of the identity doubles the exponent, so a long product folds up quickly.
(x − y)(x + y) = x2 − y2
(x2 − y2)(x2 + y2) = x4 − y4
(x4 − y4)(x4 + y4) = x8 − y8
If the chain starts at (2 + 1) with no (2 − 1) in front, supply it — it equals 1, so it changes nothing but starts the cascade.
If (2 + 1)(22 + 1)(24 + 1)(28 + 1) = 2n − 1, find n.
Multiply by (2 − 1) = 1. The chain collapses to 216 − 1, so n = 16.
The a and b can themselves be brackets. For (x + 1)2 − (2x − 3)2:
Sum = (x + 1) + (2x − 3) = 3x − 2
Difference = (x + 1) − (2x − 3) = x + 1 − 2x + 3 = 4 − x
So the factors are (3x − 2)(4 − x).
Subtracting a bracket flips every sign inside it. This single slip is the commonest error in the whole chapter. Write the minus sign out in full and change each term one at a time.
(a + b)3 = a3 + b3 + 3ab(a + b)
(a − b)3 = a3 − b3 − 3ab(a − b)
Rearranged, these give the forms you will actually use in the exam:
a3 + b3 = (a + b)3 − 3ab(a + b)
a3 − b3 = (a − b)3 + 3ab(a − b)
Sum subtracts the correction; difference adds it. One sentence covers both cube questions, and the same sentence returns in the reciprocal-powers chapter.
If a + b = 9 and ab = 14, find a3 + b3.
= 729 − 3(14)(9) = 729 − 378 = 351.
a3 + b3 = (a + b)(a2 − ab + b2)
a3 − b3 = (a − b)(a2 + ab + b2)
Use this whenever a cube sits over a trinomial in a fraction — the long bracket cancels and only the simple bracket is left.
The sum of cubes carries a minus inside the long bracket. The difference of cubes carries all pluses. It is the opposite of what most students expect, so check it every single time.
Find [(5.2)3 + (2.8)3] ÷ [(5.2)2 − 5.2×2.8 + (2.8)2].
The denominator is exactly the long bracket of a cube sum, so it cancels.
Answer = 5.2 + 2.8 = 8. No cube is ever computed.
Spot the hidden cube: 8x3 = (2x)3, 27y3 = (3y)3, 64a3 = (4a)3, 125p3 = (5p)3.
Example. If 3m + 2n = 7 and 27m3 + 8n3 = 91, find mn.
With P = 3m and Q = 2n: 91 = 343 − 21PQ, so PQ = 12. But PQ = 6mn, so mn = 2.
Divide the coefficients back out. PQ is not mn. Half the marks lost on scaled-cube questions come from reporting PQ as the final answer.
When three letters appear, three quantities matter: the total a + b + c, the sum of squares a2 + b2 + c2, and the pairwise sum ab + bc + ca. One identity links all three.
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)
Given any two of the three quantities, you can always find the third. Nothing else is needed for a large family of questions.
If a + b + c = 12 and a2 + b2 + c2 = 50, find ab + bc + ca.
144 = 50 + 2(ab + bc + ca), so 2(ab + bc + ca) = 94 and the answer is 47.
Remember the final halving — 94 is always offered as an option.
(a − b)2 + (b − c)2 + (c − a)2 = 2(a2 + b2 + c2 − ab − bc − ca)
Why this matters so much. If the right-hand side is zero, three squares add to zero. A square is never negative, so each one must be zero — which forces a = b = c.
Any question that hands you a2 + b2 + c2 = ab + bc + ca is really telling you the three letters are equal. Examiners disguise it by doubling every term, or by giving the two sides as separate equal numbers. Look for the shape, not the words.
a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)
Read it as a product of two things. Given the cubic and one factor, divide to get the other. Given both factors, multiply. No individual value of a, b or c is ever needed.
If a + b + c = 0 then the first factor is zero, so a3 + b3 + c3 = 3abc.
Always check whether the three given numbers add to zero. It is the fastest win in the chapter.
Find [(1.2)3 + (2.3)3 − (3.5)3] ÷ [1.2 × 2.3 × 3.5].
Write the top as (1.2)3 + (2.3)3 + (−3.5)3. Since 1.2 + 2.3 − 3.5 = 0, it equals 3 × 1.2 × 2.3 × (−3.5).
Cancelling against the denominator leaves −3.
Watch the sign. The subtracted cube makes the product negative while the denominator stays positive. +3 is always offered.
The condition also runs backwards. If a3 + b3 + c3 = 3abc and the total is not zero, then the second factor must vanish, which means a = b = c.
↑ Back to topWhen one equation carries two unknowns, it is almost always a hidden sum of squares. Rewrite it as squares plus a leftover and read the values straight off.
If a2 + b2 + 2a − 4b + 5 = 0, find ab.
a2 + 2a + 1 = (a + 1)2 uses 1 of the constant.
b2 − 4b + 4 = (b − 2)2 uses 4. Together they use exactly the 5.
So (a + 1)2 + (b − 2)2 = 0, giving a = −1 and b = 2, so ab = −2.
The same trick finds a minimum. Rewrite as squares plus a leftover; the leftover is the least value, because both squares can be made zero at once.
Example. 9x2 + y2 + 12x − 8y + 25 = (3x + 2)2 + (y − 4)2 + 5, so the least value is 5.
A coefficient changes the bracket. In 9x2 + 12x + 4 the bracket is (3x + 2), not (x + 2). And in a least-value question, 0 is always offered — the squares can vanish, but the leftover never does.
When four terms appear and three of them look like a square, group those three first.
x2 + 6x + 9 − y2 = (x + 3)2 − y2 = (x + 3 + y)(x + 3 − y)
The same eye finds a full cube hiding inside a longer expression: x3 − 3x2 + 3x − 1 = (x − 1)3.
Factorise the top and the bottom fully, then cancel only brackets that appear in both.
(a3 + 8) ÷ (a2 − 4) = (a + 2)(a2 − 2a + 4) ÷ (a + 2)(a − 2) = (a2 − 2a + 4) ÷ (a − 2)
You may never cancel across a plus or minus sign. Only a whole bracket that multiplies the rest can go. Cancelling a lone term out of a sum is the fastest way to a wrong option.
Six questions in the exact shapes SSC uses. Cover the working, try each one, then check.
If u + v = 9 and uv = 14, find u3 + v3 − u2 − v2.
Cubes: 729 − 378 = 351. Squares: 81 − 28 = 53.
Answer: 351 − 53 = 298. Build each piece with its own rule, then combine.
If a3 + b3 + c3 − 3abc = 108 and a2 + b2 + c2 − ab − bc − ca = 12, find a + b + c.
The identity says the first is the product of the total and the second, so 108 = (a+b+c) × 12 and the total is 9.
96 will be offered — that is 108 − 12. The identity multiplies, so recovering a factor means dividing.
If x + 1/x = 1, find x3 + 1/x3.
k3 − 3k = 1 − 3 = −2.
No real number satisfies x + 1/x = 1, because for real x that expression is at least 2 or at most −2. The identity is still valid and still fixes the value — nothing in the working assumed x was real. SSC asks these deliberately.
If p3 + q3 + 3pq = 1 with p and q positive, find p + q.
Write the 1 as −(−1)3. The equation becomes p3 + q3 + (−1)3 − 3(p)(q)(−1) = 0.
So either p + q − 1 = 0, giving p + q = 1, or p = q = −1, which the word positive rules out.
When a question adds a condition like “positive” or “x > 1”, it is almost always there to kill one of two branches. Use it rather than ignoring it.
If 4x3 + 32y3 = 432 and x + 2y = 6, find xy.
Divide by 4: x3 + 8y3 = 108. With P = x and Q = 2y, 108 = 216 − 18PQ, so PQ = 6.
PQ = 2xy, so xy = 3. Two conversions in one question, and both are easy to forget.
If 3a + 5b = 14 and 9a2 + 25b2 = 148, find the positive value of 3a − 5b.
(P − Q)2 = 2(P2 + Q2) − (P + Q)2 = 296 − 196 = 100, so the answer is 10. The product never enters.
| Identity | Use it when | Watch out for |
|---|---|---|
| (a ± b)2 = a2 ± 2ab + b2 | moving between sum/difference and squares | only the middle sign changes |
| (a+b)2 − (a−b)2 = 4ab | product wanted from two brackets | coefficients multiply into the 4 |
| (a+b)2 + (a−b)2 = 2(a2+b2) | product is unknown and not needed | remember to halve |
| a2 − b2 = (a+b)(a−b) | large numbers, decimals, chains | subtracting a bracket flips all signs |
| a3 + b3 = (a+b)3 − 3ab(a+b) | cube wanted from sum and product | sum subtracts, difference adds |
| a3 ± b3 = (a ± b)(a2 ∓ ab + b2) | cube fractions that cancel | sum → minus inside; difference → all plus |
| (a+b+c)2 = Σa2 + 2Σab | any two of total / squares / pairs | halve after subtracting |
| Σa2 = Σab ⇒ a = b = c | equality hidden in the condition | often disguised by doubling |
| a+b+c = 0 ⇒ Σa3 = 3abc | three numbers add to zero | check the sign of each cube |
| Σa3 − 3abc = (Σa)(Σa2 − Σab) | one factor given, other wanted | divide, do not subtract |
There are only about ten identities in this chapter, and every question is one of them wearing a costume — a coefficient, a shifted bracket, a decimal, or a reversed direction. Learn to strip the costume off and the chapter becomes short.
↑ Back to topThe two squares (a ± b)2, the difference of squares a2 − b2 = (a+b)(a−b), the two cube forms a3 ± b3, the squared total (a+b+c)2, and the cubic identity a3+b3+c3 − 3abc. About ten lines in total cover almost every question asked.
Use a3 + b3 = (a+b)3 − 3ab(a+b). Cube the sum, then subtract three times the product times the sum. For a difference the correction is added instead of subtracted.
The cubic identity collapses and a3 + b3 + c3 = 3abc. Whenever three numbers or three brackets in a question add to zero, this turns a long calculation into one multiplication.
That a = b = c. The condition is the same as saying the three squared differences add to zero, and a square is never negative, so each difference must be zero. Examiners often disguise it by doubling every term.
Because it is the opposite of the outside sign. A sum of cubes factorises with a minus in the long bracket, and a difference factorises with all pluses. Check it every time rather than trusting memory.
Take the matching Algebraic Formula 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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