QJEEM22Ma844a: JEE Main 2022 Mathematics - Class 11 JEEM22Ma844a Solution: The sum of the cubes of all the roots of the equation is…

${x^4} - 3{x^3} - {x^2} - {x^2} + 3x + 1 = 0$

$({x^2} - 1)({x^2} - 3x - 1) = 0$

Let the root of ${x^2} - 3x - 1 = 0$ be $\alpha$ and $\beta$ and other two roots of given equation are 1 and $-$ 1

So sum of cubes of roots

$= {1^3} + {( - 1)^3} + {\alpha ^3} + {\beta ^3}$

$= {(\alpha + \beta )^3} - 3\alpha \beta (\alpha + \beta )$

$= {(3)^3} - 3( - 1)(3)$

$= 36$