7.1 Algebraic Identities, Polynomial Factorization & Systems of Linear Equations in Two Variables

Core Algebraic Identities & Expansions

Algebraic identities are equations that remain universally true for all real values assigned to their variables. Rather than performing manual polynomial multiplication, examinees must instantly recognize patterns to factorize, cancel terms, and substitute values.

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|                         FUNDAMENTAL QUADRATIC IDENTITIES MATRIX                         |
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| Sum & Difference Squares:    (a + b)^2 = a^2 + 2ab + b^2                                |
|                              (a - b)^2 = a^2 - 2ab + b^2                                |
| Difference of Squares:       a^2 - b^2 = (a - b)(a + b)                                 |
| Symmetric Combinations:      (a + b)^2 + (a - b)^2 = 2(a^2 + b^2)                       |
|                              (a + b)^2 - (a - b)^2 = 4ab                                |
| Three-Variable Square:       (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)          |
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Cubic Identities & Factorization Structures

Cubic polynomials feature heavily in RRB examinations, testing both binomial cube expansions and the sum/difference of two cubes.

1. Binomial Cubes

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

Rearranging these identities yields the sum and difference of cubes in terms of sum/difference and product:

$$a^3 + b^3 = (a + b)^3 - 3ab(a + b)$$ $$a^3 - b^3 = (a - b)^3 + 3ab(a - b)$$

2. Factorization of Sum and Difference of Two Cubes

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

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The Three-Variable Cubic Identity & The Zero-Sum Corollary

A useful three-variable cubic identity is:

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

The Half-Sum Alternative Form

By multiplying and dividing the quadratic factor by $2$, we obtain the sum-of-squares representation:

$$a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2} \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]$$

Substituting this back into the master identity:

$$a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c) \left[ (a - b)^2 + (b - c)^2 + (c - a)^2 \right]$$

Non-Negativity & The Uniqueness of Equality

Because $(a - b)^2 \ge 0$, $(b - c)^2 \ge 0$, and $(c - a)^2 \ge 0$ for all real numbers $a, b, c$:

$$a^2 + b^2 + c^2 - ab - bc - ca \ge 0$$

Equality holds ($a^2 + b^2 + c^2 - ab - bc - ca = 0$) if and only if:

$$a - b = 0, \quad b - c = 0, \quad c - a = 0 \iff a = b = c$$

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|                        THE ZERO-SUM CONDITIONAL IDENTITY                                |
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| If:                                a + b + c = 0                                        |
|                                          │                                              |
|                                          ▼                                              |
| Then:                  a^3 + b^3 + c^3 - 3abc = (0) * (a^2 + b^2 + c^2 - ab - bc - ca)  |
|                                          │                                              |
|                                          ▼                                              |
| Result:                         a^3 + b^3 + c^3 = 3abc                                  |
+-----------------------------------------------------------------------------------------+

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Reciprocal Symmetrical Expressions ($x + \frac{1}{x} = k$)

For exercises providing $x + \frac{1}{x} = k$ or $x - \frac{1}{x} = m$ and asking for higher powers ($x^2, x^3, x^4, x^5, x^6$) can be solved by recurrence transformations. Derive each step or check it with the preceding identities.

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|                         RECIPROCAL POWER PROGRESSION LADDER                             |
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| Base Sum:                   x + 1/x = k                                                 |
|    │                                                                                    |
|    ├──> Degree 2:           x^2 + 1/x^2 = k^2 - 2                                       |
|    │                                                                                    |
|    ├──> Degree 3:           x^3 + 1/x^3 = k^3 - 3k                                      |
|    │                                                                                    |
|    ├──> Degree 4:           x^4 + 1/x^4 = (k^2 - 2)^2 - 2                               |
|    │                                                                                    |
|    ├──> Degree 5:           x^5 + 1/x^5 = (x^2 + 1/x^2)(x^3 + 1/x^3) - (x + 1/x)        |
|    │                                    = (k^2 - 2)(k^3 - 3k) - k                       |
|    │                                                                                    |
|    └──> Degree 6:           x^6 + 1/x^6 = (x^3 + 1/x^3)^2 - 2                           |
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Systematic Derivation of Power Ladders

1. Square Relation ($x^2 + \frac{1}{x^2}$):

$$\left(x + \frac{1}{x}\right)^2 = x^2 + 2(x)\left(\frac{1}{x}\right) + \frac{1}{x^2} = x^2 + \frac{1}{x^2} + 2$$ $$x^2 + \frac{1}{x^2} = k^2 - 2$$

Conversely, if $x - \frac{1}{x} = m$, then $x^2 + \frac{1}{x^2} = m^2 + 2$.

1. Cube Relation ($x^3 + \frac{1}{x^3}$):

$$\left(x + \frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3} + 3(x)\left(\frac{1}{x}\right)\left(x + \frac{1}{x}\right) = x^3 + \frac{1}{x^3} + 3k$$ $$x^3 + \frac{1}{x^3} = k^3 - 3k$$

Conversely, if $x - \frac{1}{x} = m$, then $x^3 - \frac{1}{x^3} = m^3 + 3m$.

1. Fourth Power Relation ($x^4 + \frac{1}{x^4}$): Let $P_2 = x^2 + \frac{1}{x^2} = k^2 - 2$. Then squaring $P_2$:

$$x^4 + \frac{1}{x^4} = P_2^2 - 2 = (k^2 - 2)^2 - 2$$

1. Fifth Power Relation ($x^5 + \frac{1}{x^5}$): Multiply the quadratic sum by the cubic sum:

$$\left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) = x^5 + \frac{1}{x} + x + \frac{1}{x^5} = \left(x^5 + \frac{1}{x^5}\right) + \left(x + \frac{1}{x}\right)$$ $$x^5 + \frac{1}{x^5} = \left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) - \left(x + \frac{1}{x}\right)$$

1. Sign Bridging Formula: To convert between sum and difference of reciprocals:

$$\left(x - \frac{1}{x}\right)^2 = \left(x + \frac{1}{x}\right)^2 - 4 \implies x - \frac{1}{x} = \pm \sqrt{k^2 - 4}$$ $$\left(x + \frac{1}{x}\right)^2 = \left(x - \frac{1}{x}\right)^2 + 4 \implies x + \frac{1}{x} = \pm \sqrt{m^2 + 4}$$

Special Invariant Values

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Systems of Linear Equations in Two Variables

A general system of two linear equations in two unknown variables $x$ and $y$ is expressed in standard form as:

$$a_1 x + b_1 y + c_1 = 0$$ $$a_2 x + b_2 y + c_2 = 0$$

where $a_1, b_1, c_1, a_2, b_2, c_2$ are real constants and $a_1^2 + b_1^2 \neq 0$, $a_2^2 + b_2^2 \neq 0$.

The coefficient-ratio shortcuts below require the displayed denominators to be nonzero. For zero coefficients, use elimination or the determinant test instead.

Geometric Interpretation & Consistency Criteria

Every linear equation in two variables represents a straight line on the Cartesian plane. The solution set of the system corresponds to the geometric intersection of the two lines.

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|                      LINEAR SYSTEM CLASSIFICATION & CRITERIA                            |
+-----------------------------------------------------------------------------------------+
| Ratio Condition:              Geometric State:           Algebraic Nature of Solution:  |
|                                                                                         |
|  a1/a2 != b1/b2          ---> Intersecting Lines    ---> Unique Solution (Consistent &  |
|                                                          Independent)                   |
|                                                                                         |
|  a1/a2 = b1/b2 = c1/c2   ---> Coincident Lines      ---> Infinitely Many Solutions      |
|                                                          (Consistent & Dependent)       |
|                                                                                         |
|  a1/a2 = b1/b2 != c1/c2  ---> Parallel Lines        ---> No Solution (Inconsistent)     |
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Methods of Solving Linear Systems

1. Elimination Method: Multiply one or both equations by suitable non-zero scalars so that the coefficients of one variable become equal or opposite, then add or subtract equations.
2. Substitution Method: Express one variable in terms of the other from the simpler equation and substitute into the second equation.
3. Cross-Multiplication Method (Cramer's Rule for $2 \times 2$):

$$x = \frac{b_1 c_2 - b_2 c_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{c_1 a_2 - c_2 a_1}{a_1 b_2 - a_2 b_1}$$

Notice that the denominator is the determinant $\Delta = a_1 b_2 - a_2 b_1$. A unique solution exists if and only if $\Delta \neq 0$.

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Practical Word Problems Modeled as Linear Systems

In Section Controller operations, resource allocation problems translate directly into two-variable linear models. Typical railway scenarios include:

• Rake Composition: Mixing freight wagons of different payload capacities (e.g., high-capacity BOXN coal wagons vs. covered BCN grain wagons) under total gross trailing load and maximum train length constraints.
• Passenger Ticketing & Berths: Determining passenger counts in AC 3-Tier and Sleeper coaches given total ticket revenue and total berth occupancy.
• Speed-Distance Segmentations: Dividing total route travel time between normal running speed and restricted speed over track maintenance blocks.

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Worked Examples

Worked Example 1: Three-Variable Cubic Evaluation

Problem: Given that $x + y + z = 9$ and $xy + yz + zx = 26$, determine the numerical value of $x^3 + y^3 + z^3 - 3xyz$.

• Step 1: Compute the sum of squares $x^2 + y^2 + z^2$: From the trinomial square identity:

$$(x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx)$$

Substitute the given values:

$$9^2 = x^2 + y^2 + z^2 + 2(26)$$ $$81 = x^2 + y^2 + z^2 + 52 \implies x^2 + y^2 + z^2 = 81 - 52 = 29$$

• Step 2: Apply the cubic identity:

$$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - (xy + yz + zx))$$

• Step 3: Substitute the calculated terms:

$$x^3 + y^3 + z^3 - 3xyz = 9 \times [29 - 26] = 9 \times 3 = 27$$

• Conclusion: The value of the expression is 27.

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Worked Example 2: Reciprocal Power Chain Evaluation

Problem: If $x + \frac{1}{x} = 3$, find the exact value of $x^5 + \frac{1}{x^5}$.

• Step 1: Calculate the quadratic term ($x^2 + \frac{1}{x^2}$):

$$x^2 + \frac{1}{x^2} = k^2 - 2 = 3^2 - 2 = 9 - 2 = 7$$

• Step 2: Calculate the cubic term ($x^3 + \frac{1}{x^3}$):

$$x^3 + \frac{1}{x^3} = k^3 - 3k = 3^3 - 3(3) = 27 - 9 = 18$$

• Step 3: Relate the fifth power product:

$$\left(x^2 + \frac{1}{x^2}\right)\left(x^3 + \frac{1}{x^3}\right) = \left(x^5 + \frac{1}{x^5}\right) + \left(x + \frac{1}{x}\right)$$

• Step 4: Substitute and isolate $x^5 + \frac{1}{x^5}$:

$$7 \times 18 = \left(x^5 + \frac{1}{x^5}\right) + 3$$ $$126 = \left(x^5 + \frac{1}{x^5}\right) + 3 \implies x^5 + \frac{1}{x^5} = 126 - 3 = 123$$

• Conclusion: The value of $x^5 + \frac{1}{x^5}$ is 123.

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Worked Example 3: System Dependency & Infinite Solutions

Problem: In an automated railway signaling telemetry matrix, two circuit sensor equations are given by:

$$(k - 1)x + 3y = 7$$ $$(k + 1)x + 6y = 5k - 1$$

Find the value of the parameter $k$ for which the telemetry system yields infinitely many consistent solutions.

• Step 1: State the condition for infinitely many solutions: For coincident lines, the ratio of coefficients must satisfy:

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$

Here $a_1 = k-1, b_1 = 3, c_1 = -7$; $a_2 = k+1, b_2 = 6, c_2 = -(5k-1)$.

• Step 2: Equate the first two ratios:

$$\frac{k - 1}{k + 1} = \frac{3}{6} = \frac{1}{2}$$

Cross-multiplying gives:

$$2(k - 1) = 1(k + 1) \implies 2k - 2 = k + 1 \implies k = 3$$

• Step 3: Verify the third ratio with $k = 3$:

$$\frac{c_1}{c_2} = \frac{-7}{-(5k - 1)} = \frac{7}{5(3) - 1} = \frac{7}{15 - 1} = \frac{7}{14} = \frac{1}{2}$$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2}$, $k = 3$ satisfies the complete dependency condition.

• Conclusion: The system has infinitely many solutions when $k = 3$.

The coefficient-ratio consistency test requires the denominators shown to be nonzero; otherwise use elimination or determinants. The displayed reciprocal conditions $x+1/x=1$ and $x+1/x=\sqrt{3}$ have no real solutions: their listed power identities apply only over complex numbers and are optional algebra extensions.