Question

Hello, can someone help me with the working out to this problem

Answer 1

$\text{If}~ \alpha ~\text{and}~ \beta~ \text{are the roots,}\\ \\x^2 -(\alpha + \beta)x +\alpha\beta = 0\\\\\text{For}~ x^2+3x-2=0\\\\\alpha +\beta=-\dfrac ba=-3\\ \\\alpha \beta = \dfrac ca = -2$

$\text{So,~} \alpha^3+\beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 -\alpha \beta)\\ \\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -2\alpha \beta - \alpha \beta\right]\\\\~~~~~~~~~~~~~~~~=(\alpha+\beta)\left[(\alpha +\beta)^2 -3\alpha \beta \right]\\\\~~~~~~~~~~~~~~~~=-3\left[(-3)^2-3(-2)\right]\\\\~~~~~~~~~~~~~~~~=-3(15)\\\\~~~~~~~~~~~~~~~~=-45\\\\\text{And}~~~\alpha^3 \beta^3= (\alpha \beta )^3 = (-2)^3 = -8$

$\text{Hence the equation whose roots are}~ \alpha^3 ~\text{and}~ \beta^3~ \text{is:}\\ \\ x^2-(\alpha^3 + \beta^3)x +\alpha^3 \beta^3 =0\\\\\implies x^2 -(-45)x+(-8)=0\\\\\implies x^2 +45x -8=0$