Algebra best questions for SSC CGL with notes
Hand picked best confusing questions of algebra beneficial for repeaters
How this works: the 31 questions from your Algebra vault export are regrouped below by the
identity that actually solves them, not by the order they were skipped in — grouping by underlying pattern (not
surface topic) is what makes the technique transfer to a new question in the exam. Each card uses active
recall: try the question, reveal the full working, then rate how well you knew it. Your rating schedules a
spaced-repetition review date for that question, so "Start today's session" always pulls up
exactly what's due — not everything, not randomly.
Formula cram sheet — all 10 identities on one screen
Formula notebook — every identity from your handwritten notes + the Advance Maths chapters, in one place
Transcribed from your handwritten Algebra pages and merged with the Algebra and Equations/Maxima–Minima chapters of your Advance Maths build, so every identity you've written down lives in one searchable page. Colour coding follows your own notebook: green-bordered cards are the main identity, blue is an alternate form of the same thing, gold is a technique or "how to spot it" tip, and brick-red is a trap.
Two-variable power identities
Squares
Core relations
\((a+b)^2-(a-b)^2=4ab\)
\((a+b)^2+(a-b)^2=2(a^2+b^2)\)
\(a^2+b^2=(a+b)^2-2ab=(a-b)^2+2ab\)
Cubes
Expansion
\((a+b)^3=a^3+b^3+3ab(a+b)=a^3+b^3+3a^2b+3ab^2\)
\((a-b)^3=a^3-b^3-3ab(a-b)=a^3-b^3-3a^2b+3ab^2\)
Sum / difference of cubes
\(a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)^3-3ab(a+b)=(a+b)[(a+b)^2-3ab]\)
\(a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)^3+3ab(a-b)=(a-b)[(a-b)^2+3ab]\)
Higher even powers — the halving rule
General form
\(a^n+b^n=(a^{n/2}+b^{n/2})^2-2a^{n/2}b^{n/2}\)
Applied
\(a^4+b^4=(a^2+b^2)^2-2a^2b^2\)
\(a^6+b^6=(a^2+b^2)^3-3a^2b^2(a^2+b^2)=(a^3+b^3)^2-2a^3b^3\)
\(a^8+b^8=(a^4+b^4)^2-2a^4b^4\)
Odd powers — the cross-multiply rule
Sum of odd powers
\(a^3+b^3=(a+b)^3-3ab(a+b)\)
\(a^5+b^5=(a^2+b^2)(a^3+b^3)-a^2b^2(a+b)\)
\(a^7+b^7=(a^3+b^3)(a^4+b^4)-a^3b^3(a+b)\)
Difference of odd powers
\(a^3-b^3=(a-b)^3+3ab(a-b)\)
\(a^5-b^5=(a^2-b^2)(a^3+b^3)+a^2b^2(a-b)\)
\(a^7-b^7=(a^3-b^3)(a^4+b^4)+a^3b^3(a-b)\)
Which sign does the correction take?
Both brackets plus → always subtract the overlap
Minus on the smaller exponent → add the overlap
Minus on the larger exponent → subtract the overlap
So \(a^5-b^5\) has two valid forms with opposite-sign corrections:
\(a^5-b^5=(a^2-b^2)(a^3+b^3)+a^2b^2(a-b)\) — minus on the 2
\(a^5-b^5=(a^2+b^2)(a^3-b^3)-a^2b^2(a-b)\) — minus on the 3
If unsure, expand both brackets for a=2, b=1 and see which way the correction goes.
\(a^5-b^5=(a^2-b^2)(a^3+b^3)+a^2b^2(a-b)\) — minus on the 2
\(a^5-b^5=(a^2+b^2)(a^3-b^3)-a^2b^2(a-b)\) — minus on the 3
If unsure, expand both brackets for a=2, b=1 and see which way the correction goes.
Differences and factorisation
Difference of powers
\(a^2-b^2=(a-b)(a+b)\)
\(a^4-b^4=(a^2-b^2)(a^2+b^2)\)
\(a^6-b^6=(a^3-b^3)(a^3+b^3)=(a^2-b^2)^3+3a^2b^2(a^2-b^2)\)
The \(a^n-1\) family
\(a^n-1=(a-1)(a^{n-1}+a^{n-2}+\cdots+a+1)\)
\(a^4-1=(a-1)(a^3+a^2+a+1)\)
Symmetric "fourth-power-family" identity
General form
\(x^n+y^n+x^{n/2}y^{n/2}=(x^{n/2}+y^{n/2}+(xy)^{n/4})(x^{n/2}+y^{n/2}-(xy)^{n/4})\)
Same identity, written for \(n=4\): \(x^4+y^4+x^2y^2=(x^2+y^2)^2-x^2y^2\). This is the one behind "if \(x^2-x+1=0\), find \(x^4+x^2+1\)"-style questions.
Three (and more)-variable identities
The workhorse
\(a^3+b^3+c^3-3abc\)
\(a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\)
\(\phantom{a^3+b^3+c^3-3abc}=(a+b+c)\times\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]\)
The two conclusions examiners want
If \(a+b+c=0\) → \(a^3+b^3+c^3=3abc\)
If \(a^3+b^3+c^3=3abc\) and \(a+b+c\ne0\) → \(a=b=c\)
Squares and cubes of a multi-term sum
Expansions
\((a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\)
\((a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(c+a)\)
\(\phantom{(a+b+c)^3}=a^3+b^3+c^3+3a^2(b+c)+3b^2(c+a)+3c^2(a+b)+6abc\)
\((a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)\)
Differences of differences
\((a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)\)
\((a-b)^2+(b-c)^2+(c-a)^2=2(a^2+b^2+c^2)-2(ab+bc+ca)\)
Three-term cube-of-sums
\((x+y)^3+(y+z)^3+(z+x)^3=2(x^3+y^3+z^3)+3(x+y)(y+z)(z+x)-6xyz\)
Two forms of the same thing
\(a^2+b^2+c^2-ab-bc-ca\)
\(=\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]\)
\(=(a+b+c)^2-3(ab+bc+ca)\)
Sum-of-squares = 0
If \((m-5)^2+(n-7)^2+(p-6)^2=0\) → \(m=5,\ n=7,\ p=6\)
If \(a^2+b^2+c^2=ab+bc+ca\) → \(a=b=c\)
If \(a+b+c=k\sqrt3\) and \(a^2+b^2+c^2=k^2\) → \(a=b=c\)
Real squares can't be negative, so each bracket must be zero on its own. Any question that opens with a scattered quadratic and asks for a single value is usually this.
Useful expansions
Expansion worth remembering
\((a+b+c)(ab+bc+ca)=a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+3abc\)
"Vector-sum-of-squares" identity
\((ax+by)^2+(bx-ay)^2=(a^2+b^2)(x^2+y^2)\)
The one behind Unit 8's "ax+by=6, bx−ay=2, a²+b²=4" style question.
Two-bracket factoring
\(a+b+ab+1=(a+1)(b+1)\)
\(ab-a-b+1=(a-1)(b-1)\)
x + 1/x — the biggest scoring topic
The ladder
Going up needs no sign check
\(x+\tfrac1x=a\ \Rightarrow\ x^2+\tfrac1{x^2}=a^2-2\)
\(x-\tfrac1x=a\ \Rightarrow\ x^2+\tfrac1{x^2}=a^2+2\)
Going down needs a ± check
\(x^2+\tfrac1{x^2}=k\ \Rightarrow\ x+\tfrac1x=\pm\sqrt{k+2}\ ,\ \ x-\tfrac1x=\pm\sqrt{k-2}\)
\(x+\tfrac1x=k\ \Rightarrow\ x-\tfrac1x=\pm\sqrt{k^2-4}\)
\(x-\tfrac1x=k\ \Rightarrow\ x+\tfrac1x=\pm\sqrt{k^2+4}\)
A square root has two signs. \(x^4+1/x^4=62\) gives \(x^2+1/x^2=8\), then \(x-1/x=\pm\sqrt6\) — check what the question actually needs before defaulting to the positive root.
Cubes and climbing to any power
Cubes
\(x+\tfrac1x=m\ \Rightarrow\ x^3+\tfrac1{x^3}=m^3-3m\)
\(x-\tfrac1x=m\ \Rightarrow\ x^3-\tfrac1{x^3}=m^3+3m\)
The chaining rule
\((x^a+\tfrac1{x^a})(x^b+\tfrac1{x^b})=x^{a+b}+\tfrac1{x^{a+b}}+x^{a-b}+\tfrac1{x^{a-b}}\)
\(x^5+\tfrac1{x^5}=(x^2+\tfrac1{x^2})(x^3+\tfrac1{x^3})-(x+\tfrac1x)\)
\(x^5-\tfrac1{x^5}=(x^2-\tfrac1{x^2})(x^3+\tfrac1{x^3})+(x-\tfrac1x)=(x^2+\tfrac1{x^2})(x^3-\tfrac1{x^3})-(x-\tfrac1x)\)
\(x^7+\tfrac1{x^7}=(x^3+\tfrac1{x^3})(x^4+\tfrac1{x^4})-(x+\tfrac1x)\)
\(x^7-\tfrac1{x^7}=(x^3-\tfrac1{x^3})(x^4+\tfrac1{x^4})+(x-\tfrac1x)\)
Multiply two rungs of the ladder, then subtract (or add) the leftover rung. With \(x+1/x=3\): \(x^5+1/x^5=123\) and \(x^7+1/x^7=843\).
Special values — memorise, never derive
Fixed points
\(x+\tfrac1x=2\ \Rightarrow\ x=1\)
\(x+\tfrac1x=-2\ \Rightarrow\ x=-1\)
\(x+\tfrac1x=1\ \Rightarrow\ x^2-x+1=0\ \Rightarrow\ x^3=-1\)
\(x+\tfrac1x=-1\ \Rightarrow\ x^2+x+1=0\ \Rightarrow\ x^3=1\)
\(x+\tfrac1x=\pm\sqrt3\ \Rightarrow\ x^6=-1\ ,\ \ x^3+\tfrac1{x^3}=0\)
Once \(x^3=\pm1\), any power reduces instantly. \(x+1/x=1\) gives \(x^{29}+1/x^{29}=-1\).
Same rule in ratio form
\(\tfrac pq+\tfrac qp=2\ \Rightarrow\ p=q\)
\(\tfrac pq+\tfrac qp=-2\ \Rightarrow\ p=-q\)
\(\tfrac pq+\tfrac qp=1\ \Rightarrow\ (\tfrac pq)^3=-1\)
\(\tfrac pq+\tfrac qp=-1\ \Rightarrow\ (\tfrac pq)^3=1\)
So \(k/4+4/k=1\) gives \((k/4)^3=-1\), hence \(k^3=-64\).
Reciprocal-pair sum = 1
If \(xy=1\), then \(\dfrac1{x+1}+\dfrac1{y+1}=1\)
Exactly the shortcut used in Unit 4's Q23 — works for any x, y as long as their product is 1.
Getting into x + 1/x shape
Four common disguises
- Quadratic given: \(x^2-5x-1=0\) → divide through by x → \(x-1/x=5\).
- Shifted variable: \(x+1/(x-29)=27\) → subtract 29 from both sides so the bracket matches.
- Mismatched coefficients: \(2x+1/(3x)=8\) and you need \(9x^2+1/(4x^2)\) — multiply the given equation by whatever factor makes the coefficients line up, squaring before or after as needed. (This is exactly the trick behind Unit 2's Q16.)
- Big fraction: divide numerator and denominator by the middle power of x. For \((x^4+4x^3+5x^2+4x+1)/(x^2+1)\), divide by \(x^2\).
Value-putting
When variables outnumber equations
If the question gives fewer equations than variables, the answer is independent of the actual values — so pick easy numbers that satisfy the given condition.
Given \(x+y+z=0\), just take \(x=0,\ y=1,\ z=-1\) and evaluate directly. This is exactly how Unit 10's Q30 gets solved.
Linear systems
Consistency of two lines
| Ratio test | Meaning |
|---|---|
| \(a_1/a_2\ne b_1/b_2\) | Intersecting → unique solution → consistent |
| \(a_1/a_2=b_1/b_2\ne c_1/c_2\) | Parallel → no solution → inconsistent |
| \(a_1/a_2=b_1/b_2=c_1/c_2\) | Coincident → infinite solutions → consistent |
Counting solutions
- 1 variable, 1 equation → 1 value. n variables with n independent equations → all n values are findable.
- Three variables: eliminate one using a pair of equations to reduce to a two-variable system.
- A three-variable linear equation represents a plane, not a line.
Quadratic equations
Roots and reconstruction
\(ax^2+bx+c=0\)
\(\alpha+\beta=-b/a\ ,\ \ \alpha\beta=c/a\)
\(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\)
To rebuild: \(x^2-(\text{sum})x+(\text{product})=0\)
Nature of roots — \(D=b^2-4ac\)
| Discriminant | Roots |
|---|---|
| D>0, perfect square | Real, unequal, rational |
| D>0, not a perfect square | Real, unequal, irrational |
| D=0 | Real and equal |
| D<0 | Imaginary, unequal |
Irrational and imaginary roots always come in conjugate pairs — if one root is \(2+\sqrt3\), the other is \(2-\sqrt3\).
Reading signs off the equation
Only the middle term negative → both roots positive
Only the constant term negative → roots have opposite signs
Sign of the whole expression
\(ax^2+bx+c=a(x-\alpha)(x-\beta)\)
x outside the roots → same sign as a
x between the roots → opposite sign to a
Modulus equations — number of solutions
\(x^2+5|x|+6=0\) → 0 solutions (all terms positive)
\(x^2-5|x|+6=0\) → 4 solutions (middle term negative)
\(x^2+5|x|-6=0\) → 2 solutions (constant term negative)
Cubic and higher-degree equations
Cubic relations
\(ax^3+bx^2+cx+d=0\) with roots \(p,q,r\):
\(p+q+r=-b/a\ ,\ \ pq+qr+rp=c/a\ ,\ \ pqr=-d/a\)
Find one root by hit and trial, then use synthetic division to drop to a quadratic. A degree-n equation has at most n roots.
Maxima and minima
The quadratic vertex
For \(ax^2+bx+c\), the turning point sits at \(x=-b/2a\), value there \(=(4ac-b^2)/4a\)
\(a>0\) → that value is the minimum (max is \(\infty\))
\(a<0\) → that value is the maximum (min is \(-\infty\))
Or complete the square: \(x^2-4x+5=(x-2)^2+1\), so the minimum is 1. This is also the method behind the "minimize \(9a^2+4b^2+49c^2+16\)" style example in the notes — complete the square in each variable separately and read off the vertex.
Reciprocal minimums
\(m>0\) → \(m+\tfrac1m\) has minimum 2
Any nonzero real \(x\), \(n\) even → \(x^n+\tfrac1{x^n}\ge2\)
\(a,b,c>0\) → \((a+b+c)(\tfrac1a+\tfrac1b+\tfrac1c)\) has minimum 9
AM ≥ GM
\(\dfrac{a+b+c+d}{4}\ge\sqrt[4]{abcd}\)
With a fixed sum, the product is largest when all the terms are equal — that equality case answers almost every "find the maximum product" question.
Shifting the expression
For something like \(a^2+\dfrac1{a^2-2}\) with \(a>2\), add and subtract the offending constant so you can apply \(m+1/m\ge2\) to the shifted term, then add back what you took out.