Given Sum and Product, Compute Any Power Sum — Without Ever Finding a and b

JEE Mains will give you eighty seconds per question. JEE Advanced even less per multi-step. So when a question says "If a + b = 5 and ab = 6, find a^3 + b^3", you do not have time to set up t^2 - 5t + 6 = 0, factor it, find a = 2 and b = 3, cube them, and add. That is forty seconds you don't have.

The recognition rule is this: inputs of the form (sum, product) deserve outputs computed by symmetric identities, full stop. The roots themselves are a distraction. This article is the exam-focused lookup card for that recognition.

(For the underlying intuition — why symmetric inputs deserve symmetric tools — see Symmetric Expressions: Think Identities, Not Brute Force. For a deep zoom on the workhorse a^2 + b^2 case alone, see x^2 + y^2 as (x+y)^2 - 2xy. This article is the exam-room reference card that ties them together — a recognition tool, not a derivation.)

The lookup card — memorise these five lines

Throughout, s = a+b and p = ab.

A line-by-line tour:

• a^2 + b^2 = s^2 - 2p. From (a+b)^2 = a^2 + 2ab + b^2, subtract 2ab. The most-used line on the card. Why subtracting 2p is the right move: squaring a sum produces a "cross-term tax" of 2ab. Subtracting it isolates the pure sum of squares.
• a^2 - b^2 = (a-b)(a+b) = s\sqrt{s^2 - 4p}. Use (a-b)^2 = (a+b)^2 - 4ab = s^2 - 4p, so |a - b| = \sqrt{s^2 - 4p}. The sign is determined by which of a, b is bigger — the symmetric inputs alone cannot fix the order. (The expression a^2 - b^2 itself is not symmetric — a giveaway that you need extra information.)
• a^3 + b^3 = s(s^2 - 3p). From (a+b)^3 = a^3 + b^3 + 3ab(a+b), isolate a^3 + b^3 = s^3 - 3ps = s(s^2 - 3p). The factored form is faster to plug numbers into.
• a^3 - b^3 = \sqrt{s^2 - 4p} \cdot (s^2 - p). From the standard identity a^3 - b^3 = (a - b)(a^2 + ab + b^2), where a^2 + ab + b^2 = (a^2 + b^2) + ab = (s^2 - 2p) + p = s^2 - p, and |a - b| = \sqrt{s^2 - 4p}.
• a^4 + b^4 = (s^2 - 2p)^2 - 2p^2. Compute a^2 + b^2 = s^2 - 2p first, then bootstrap: a^4 + b^4 = (a^2 + b^2)^2 - 2(ab)^2. Why this works: (a^2 + b^2)^2 expands to a^4 + 2a^2 b^2 + b^4, and a^2 b^2 = p^2. Subtract 2p^2 to recover the pure a^4 + b^4. You are applying the degree-2 trick at degree 4.

Why solving for a and b would lose marks and time: even if the roots are nice integers, the substitution route is \sim 5× faster. And the moment the roots become irrational — discriminant s^2 - 4p is not a perfect square, e.g. s = 3, p = 1 gives roots (3 \pm \sqrt{5})/2 — cubing surds and adding them is brutal. The symmetric route doesn't notice or care that the roots are ugly.

Worked example 1 — straight substitution

Worked example 2 — the a + 1/a disguise

Worked example 3 — Newton's recurrence

How to recognise the pattern in an exam

Train your eye for these triggers. If any one of them lights up, reach for the card.

• The problem explicitly gives a + b and ab (or a - b and ab).
• The problem features a + 1/a or a - 1/a — hidden p = 1.
• The problem mentions roots of a quadratic t^2 + bt + c = 0. By Vieta's relations, the roots \alpha, \beta satisfy \alpha + \beta = -b and \alpha \beta = c. You are handed sum and product on a plate.
• The problem asks for a^k + b^k, a^k - b^k, or any expression that does not change when you swap a and b.

When the trigger fires, do not solve the quadratic. Read s and p off, look up the relevant line of the card, substitute. Answer in twenty seconds.

Common confusions

• "a^2 - b^2 should be computable from s and p alone." It is not. The expression a^2 - b^2 = (a-b)(a+b) is antisymmetric — it flips sign when you swap a and b. Symmetric inputs (s and p) cannot determine the sign of an antisymmetric output. You need the extra bit "which of a, b is bigger" to fix the sign of \sqrt{s^2 - 4p}. This is also why the lookup card writes \sqrt{s^2 - 4p} as a magnitude.
• "The discriminant s^2 - 4p might be negative." If s, p come from real numbers a, b, then s^2 - 4p = (a - b)^2 \geq 0 always. So the square root is real. (If a problem hands you arbitrary s, p without claiming real roots, the square root could be complex — but that is a different setup.)
• "a^3 + b^3 = s^3 - 3p — I'll just memorise that." Wrong by a factor. The correct line is a^3 + b^3 = s^3 - 3ps, which factors as s(s^2 - 3p). The dimension check helps: a^3 + b^3 is degree 3 in a, b, so every term must be degree 3 in s, p (where s counts as degree 1 and p as degree 2). s^3 is degree 3; ps is degree 3; bare p would be degree 2 and is wrong.
• "For a + 1/a, the product ab is a — not 1." No. With b = 1/a, the product is a \cdot (1/a) = 1, not a. The whole point of the substitution is that b's value depends on a such that the product collapses to a constant.

Going deeper

Newton's identities — the general principle

Newton's identities are the systematic version of the recurrence. For a polynomial with roots r_1, r_2, \ldots, r_n, define the power sums P_k = r_1^k + r_2^k + \cdots + r_n^k and the elementary symmetric polynomials e_1, e_2, \ldots, e_n (the coefficients of the polynomial up to sign, by Vieta's formulas). Newton's identities say:

For two variables, this collapses to P_k = s P_{k-1} - p P_{k-2} — exactly what Example 3 used. The same recurrence, applied with three or more roots, lets you compute power sums of cubic, quartic, and higher equations without ever solving them. This is the engine behind a large fraction of JEE Advanced and Olympiad algebra.

The fundamental theorem of symmetric polynomials

The deeper reason any a^k + b^k can be written using only s and p is the fundamental theorem of symmetric polynomials: every polynomial expression in a, b that is unchanged under swapping them ("symmetric") can be uniquely written as a polynomial in e_1 = s and e_2 = p. Power sums P_k are symmetric, so they qualify. The lookup card is the first few cases of this theorem made explicit. The theorem itself, generalised, drives Galois theory and the proof that the general quintic has no radical solution.

References

• Newton's identities — Wikipedia
• Elementary symmetric polynomial — Wikipedia
• Vieta's formulas — Wikipedia
• Power sum — Wikipedia
• Symmetric polynomial — Wikipedia