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CDS Mathematics · Algebraic Identities and Simplification

Reciprocal Sums x ± 1/x

Once x + 1/x or x − 1/x is known, every higher power sum like x² + 1/x², x³ + 1/x³ or x⁴ − 1/x⁴ follows by squaring or cubing.

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Why this matters

Eleven PYQs, almost every paper has one. The work is a short ladder — square to go up two powers, cube to go up three, take a square root to come down — and the only real danger is the sign of x − 1/x.

Concept 1 of 2: Squaring up and down the ladder

Squaring x+1xx + \dfrac1x gives x2+1x2x^2 + \dfrac{1}{x^2} plus 22, because the cross term x⋅1xx\cdot\dfrac1x is 11. So each squaring moves two powers up, and a square root moves back down.

Definition

  • (x+1x)2=x2+1x2+2\left(x + \dfrac1x\right)^2 = x^2 + \dfrac{1}{x^2} + 2 and (x−1x)2=x2+1x2−2\left(x - \dfrac1x\right)^2 = x^2 + \dfrac{1}{x^2} - 2.
  • So (x+1x)2−(x−1x)2=4\left(x + \dfrac1x\right)^2 - \left(x - \dfrac1x\right)^2 = 4.
  • x2−1x2=(x−1x)(x+1x)x^2 - \dfrac{1}{x^2} = \left(x - \dfrac1x\right)\left(x + \dfrac1x\right) and x4+1x4=(x2+1x2)2−2x^4 + \dfrac{1}{x^4} = \left(x^2 + \dfrac{1}{x^2}\right)^2 - 2.
  • An expression like xx2+kx+1\dfrac{x}{x^2 + kx + 1}: divide top and bottom by xx to get 1x+1x+k\dfrac{1}{x + \frac1x + k}.
  • xy−yx\sqrt{\dfrac xy} - \sqrt{\dfrac yx} is r−1rr - \dfrac1r with r=xyr = \sqrt{\dfrac xy}.

Squaring a reciprocal sum

(x±1x)2=x2+1x2±2\left(x \pm \tfrac1x\right)^2 = x^2 + \tfrac{1}{x^2} \pm 2

Worked example

If x+1x=3x + \dfrac1x = 3, find x2+1x2x^2 + \dfrac{1}{x^2}, x4+1x4x^4 + \dfrac{1}{x^4} and x−1xx - \dfrac1x.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2025 · CDS (II) 2025 — Elementary Mathematics · Q26Easy

Example 1 · Algebraic Identities and Simplification · Reciprocal Sums x ± 1/x

If x−1x=2x-\frac{1}{x}=2, x>0x>0; then what is x2−1x2x^2-\frac{1}{x^2} equal to?

x − 1/x has two signs

Coming down from x2+1x2x^2 + \dfrac{1}{x^2}, x−1xx - \dfrac1x is ±\pm a square root, and x>0x > 0 does not decide the sign (x=2x = 2 and x=12x = \dfrac12 both have x>0x > 0). If the options list only one sign, that is the intended value; if both appear, look for a condition like x>1x > 1.
Drill 6 more on squaring up and down the ladder

Concept 2 of 2: Cubes and higher powers

Cubing x+1xx + \dfrac1x gives x3+1x3x^3 + \dfrac{1}{x^3} plus three copies of x+1xx + \dfrac1x itself. Subtract them and the cube is known.

Definition

With u=x+1xu = x + \dfrac1x and v=x−1xv = x - \dfrac1x:

  • x3+1x3=u3−3ux^3 + \dfrac{1}{x^3} = u^3 - 3u
  • x3−1x3=v3+3vx^3 - \dfrac{1}{x^3} = v^3 + 3v
  • x6+1x6=w3−3wx^6 + \dfrac{1}{x^6} = w^3 - 3w with w=x2+1x2w = x^2 + \dfrac{1}{x^2}, so x6+1x6=w(w2−3)x^6 + \dfrac{1}{x^6} = w(w^2 - 3).
  • a6−1a3\dfrac{a^6 - 1}{a^3} is just a3−1a3a^3 - \dfrac{1}{a^3}, and a2−1a\dfrac{a^2 - 1}{a} is a−1aa - \dfrac1a.

Cubes of a reciprocal sum

x3+1x3=u3−3u,x3−1x3=v3+3vx^3 + \tfrac{1}{x^3} = u^3 - 3u, \qquad x^3 - \tfrac{1}{x^3} = v^3 + 3v

Worked example

If x+1x=3x + \dfrac1x = 3, find x3+1x3x^3 + \dfrac{1}{x^3}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (I) 2018 — Elementary Mathematics · Q9Moderate

Example 2 · Algebraic Identities and Simplification · Reciprocal Sums x ± 1/x

If a2−1a=5\frac{a^2 - 1}{a} = 5, then what is the value of a6−1a3\frac{a^6 - 1}{a^3}?

Plus 3v for the difference, minus 3u for the sum

(x−1x)3=x3−1x3−3(x−1x)\left(x - \dfrac1x\right)^3 = x^3 - \dfrac{1}{x^3} - 3\left(x - \dfrac1x\right), so the difference of cubes ADDS 3v3v. Using −3v-3v by analogy with the sum gives a value that is usually one of the options.
Drill 3 more on cubes and higher powers

Summary — formulas & gotchas at a glance

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Formulas (2)

  • Squaring up and down the ladder

    Squaring a reciprocal sum

    (x±1x)2=x2+1x2±2\left(x \pm \tfrac1x\right)^2 = x^2 + \tfrac{1}{x^2} \pm 2
  • Cubes and higher powers

    Cubes of a reciprocal sum

    x3+1x3=u3−3u,x3−1x3=v3+3vx^3 + \tfrac{1}{x^3} = u^3 - 3u, \qquad x^3 - \tfrac{1}{x^3} = v^3 + 3v

Watch out for (2)

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