Algebra — SSC CGL Quantitative Aptitude

SSC does not test whether you can do algebra; it tests whether you recognise which of about ten identities a question is built on. The same families repeat every cycle: the $x+\frac{1}{x}$ chain, the three-variable cube identity, surd rationalisation, and "value-putting". If geometry is a theorem-recognition game, algebra is an identity-recognition game — and the identity list is even shorter.

1. What SSC actually asks

Tier 1: 2–3 questions · Tier 2: 3–5 questions. The recurring families:

1. $x+\frac{1}{x}$ chains — given $x+\frac{1}{x}=k$, find $x^2+\frac{1}{x^2}$, $x^3+\frac{1}{x^3}$, $x^4+\frac{1}{x^4}$…
2. Three-variable cubes — $a^3+b^3+c^3-3abc$ when $a+b+c=0$ or symmetric values are given.
3. Square-sum tricks — $a^2+b^2+c^2$ from $a+b+c$ and $ab+bc+ca$.
4. Surds — rationalise $\frac{1}{\sqrt{7}-\sqrt{6}}$, simplify telescoping surd sums, compare surds.
5. Value-putting / symmetry — monstrous-looking expressions that collapse for a smart substitution.
6. Linear equations & graphs (Tier 2) — solution conditions, intersection points.

2. The core identity kit

$(a+b{)}^2=a^2+2ab+b^2(a-b{)}^2=a^2-2ab+b^2$ $a^2-b^2=(a+b)(a-b)$ $(a+b{)}^3=a^3+b^3+3ab(a+b)(a-b{)}^3=a^3-b^3-3ab(a-b)$ $a^3+b^3=(a+b)(a^2-ab+b^2)a^3-b^3=(a-b)(a^2+ab+b^2)$

The three-variable monsters (memorise both directions):

$(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Two instant corollaries SSC loves:

• If $a+b+c=0$ then $a^3+b^3+c^3=3abc$. (Spot hidden zero-sums: $(x-4),(x-5),(9-2x)$ sum to 0!)
• $a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\right]$ — zero iff $a=b=c$.

3. The $x+\frac{1}{x}$ chain — SSC's favourite machine

From $x+\frac{1}{x}=k$, everything follows by squaring and cubing:

For the minus version $x-\frac{1}{x}=m$: $x^2+\frac{1}{x^2}=m^2+2$ and $x^3-\frac{1}{x^3}=m^3+3m$.

Special values to know cold:

• $x+\frac{1}{x}=2\Rightarrow x=1$ (so any $x^n+\frac{1}{x^n}=2$).
• $x+\frac{1}{x}=-2\Rightarrow x=-1$.
• $x+\frac{1}{x}=1\Rightarrow x^3=-1$ → powers cycle with period 6. E.g. $x^{18}+\frac{1}{x^{18}}=2$.
• $x+\frac{1}{x}=\sqrt{3}\Rightarrow x^6=-1$ → period 12.

When the given value is 1, √3, or −1, do not grind the chain — jump to the cube/cyclic fact.

4. Surds & rationalisation

Rationalise with the conjugate:

$\frac{1}{\sqrt{a}-\sqrt{b}}=\frac{\sqrt{a}+\sqrt{b}}{a-b}$

Telescoping sums (a Tier-2 staple): each term of $\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\cdots +\frac{1}{\sqrt{8}+3}$ rationalises to $\sqrt{n+1}-\sqrt{n}$; the sum collapses to $3-1=2$.

Nested surds: $\sqrt{a\pm 2\sqrt{b}}=\sqrt{x}\pm \sqrt{y}$ where $x+y=a$, $xy=b$. E.g. $\sqrt{7+2\sqrt{12}}=\sqrt{4}+\sqrt{3}=2+\sqrt{3}$.

Comparing surds: raise to a common power: $\sqrt{2}$ vs $\sqrt[3]{3}$ → compare $2^3=8$ vs $3^2=9$ → $\sqrt[3]{3}$ is bigger.

5. Componendo–dividendo and ratio algebra

If $\frac{a}{b}=\frac{c}{d}$, then $\frac{a+b}{a-b}=\frac{c+d}{c-d}$ (componendo–dividendo).

Classic ask: "If $\frac{x}{y}=\frac{4}{5}$, find $\frac{4x+5y}{4x-5y}$." Substitute $x=4t,y=5t$: $\frac{16t+25t}{16t-25t}=-\frac{41}{9}$. Substitution beats manipulation — pick the smallest consistent values and evaluate.

6. Linear equations & graphs (Tier 2)

For $a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$:

Also on tap: $ax+by=c$ meets the axes at $\left(\frac{c}{a},0\right)$ and $\left(0,\frac{c}{b}\right)$; the triangle formed with the axes has area $\frac{1}{2}\left|\frac{c}{a}\right|\left|\frac{c}{b}\right|$.

7. Solved PYQ-style examples

Q1. If $x+\frac{1}{x}=5$, find $x^3+\frac{1}{x^3}$. Solution. $k^3-3k=125-15=110$.

Q2. If $a+b+c=9$ and $ab+bc+ca=26$, find $a^2+b^2+c^2$. Solution. $(a+b+c{)}^2-2(ab+bc+ca)=81-52=29$.

Q3. Evaluate $(x-4{)}^3+(x-9{)}^3+(13-2x{)}^3$ at $x=5$. Solution. First check the bases: $(x-4)+(x-9)+(13-2x)=0$ — a hidden zero-sum! So the expression $=3(x-4)(x-9)(13-2x)$. At $x=5$: $3(1)(-4)(3)=-36$. (Whenever three cubes are summed, test the bases for a zero-sum before expanding anything.)

Q4. If $x=7+4\sqrt{3}$, find $\sqrt{x}+\frac{1}{\sqrt{x}}$. Solution. $x=(2+\sqrt{3}{)}^2$, so $\sqrt{x}=2+\sqrt{3}$ and $\frac{1}{\sqrt{x}}=2-\sqrt{3}$. Sum $=4$. (Also: $x+\frac{1}{x}=14$, and $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt{14+2}=4$.)

Q5. If $a^2+b^2+c^2=ab+bc+ca$, find $\frac{a+c}{b}$. Solution. The condition forces $a=b=c$ (sum-of-squares corollary) → $\frac{a+c}{b}=2$.

Q6 (Tier 2). For what value of $k$ do $3x+4y=7$ and $kx+8y=10$ have no solution? Solution. Need $\frac{3}{k}=\frac{4}{8}\ne \frac{7}{10}$ → $k=6$ (and $\frac{1}{2}\ne \frac{7}{10}$ ✓).

8. Training protocol

Algebra is the topic where formula recall converts directly into marks. Write the identity kit (Sections 2–3) daily for a week. Then drill PYQs in identity families, asking one question each time: which identity is this dressed in? When stuck for 20 seconds in the exam, try substitution — pick a legal value, evaluate, and match options. Substitution rescues at least one algebra question per paper.