If α, β and γ are the zeros of $x^3-6x^2+7x-2$, then α² + β² + γ² equals:

For a cubic the zeros satisfy α + β + γ = 6 and αβ + βγ + γα = 7. Then α² + β² + γ² = $(\alpha +\beta +\gamma {)}^2-2$(αβ + βγ + γα) = 36 − 14 = 22. (36 stops at the square of the sum; 29 subtracts the pair-sum once instead of twice; 50 adds 14 instead of subtracting it.) The constant term −2 plays no part — it fixes the product αβγ, which this expression never uses.