JEE Main 2022 — 26 June, Morning Shift. Previous Year Question.
Problem
The sum of the cubes of all the roots of the equation
$$x^4 – 3x^3 – 2x^2 + 3x + 1 = 0$$
is?
Key insight. There’s no standard formula for the sum of cubes of the roots of a quartic the way there is for a quadratic — so instead of hunting for one, the equation itself can be rearranged and factored into two quadratics whose roots are easy to combine using $a^3+b^3=(a+b)^3-3ab(a+b)$.
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Approach
Group the quartic’s terms so that it reveals a perfect square minus a multiple of another expression, which factors into two quadratics. One quadratic gives two roots outright; the other gives a sum and product of roots that can be plugged directly into the cube-sum identity — without ever solving for the roots individually.
Solution
Step 1 — Regroup the quartic
$$x^4 – 3x^3 – 2x^2 + 3x + 1 = (x^4 – 2x^2 + 1) – (3x^3 – 3x) = (x^2-1)^2 – 3x(x^2-1)$$
Step 2 — Factor out the common term
$$(x^2-1)^2 – 3x(x^2-1) = (x^2-1)\big[(x^2-1) – 3x\big] = (x^2-1)(x^2-3x-1)$$
Step 3 — Solve each quadratic factor
From $x^2-1=0$: two roots $\alpha = 1,\ \beta = -1$.
From $x^2-3x-1=0$: this doesn’t factor nicely, but by Vieta’s formulas its roots $\gamma,\delta$ satisfy:
$$\gamma+\delta = 3, \qquad \gamma\delta = -1$$
Step 4 — Sum the cubes of the first two roots
$$\alpha^3+\beta^3 = 1^3+(-1)^3 = 0$$
Step 5 — Sum the cubes of the other two roots using the identity
$$\gamma^3+\delta^3 = (\gamma+\delta)^3 – 3\gamma\delta(\gamma+\delta)$$
$$= 3^3 – 3(-1)(3) = 27 + 9 = 36$$
Step 6 — Combine all four
$$\alpha^3+\beta^3+\gamma^3+\delta^3 = 0 + 36 = 36$$
Answer
$$36$$
Common mistakes
- Trying to solve $x^2-3x-1=0$ using the quadratic formula and cubing the surds directly. This works but turns into messy arithmetic with $\sqrt{13}$; using the identity $a^3+b^3=(a+b)^3-3ab(a+b)$ with just the sum and product avoids surds entirely.
- Missing the regrouping step. Without spotting that $x^4-2x^2+1$ is $(x^2-1)^2$, the quartic looks unfactorable and forces a much longer numerical approach.
Practise next
- Find the sum of the cubes of all roots of $x^4 – 5x^3 + 5x^2 + 5x – 1 = 0$, checking first whether a similar regrouping into two quadratics is possible.
Show answer
$35$. No regrouping into two quadratics is available here, so use Newton’s identities directly on $e_1=5$, $e_2=5$, $e_3=-5$, $e_4=-1$.
$p_1=e_1=5$; $p_2=e_1p_1-2e_2=25-10=15$; and
$$p_3=e_1p_2-e_2p_1+3e_3=75-25-15=35.$$
Checking the four roots numerically and cubing them gives $35$ as well. Newton’s identities never need the roots themselves — which is the point when the quartic does not factor.
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