Question

if a+b+c=1 and ab+bc+ca=20 find the value of a^3+b^3+c^3-3abc

Answer 1

${(a + b + c)}^{2} = ({a}^{2} + {b}^{2} + {c}^{2}) + (2 ab + 2bc + 2ca) \\ 1 = {a}^{2} + {b}^{2} + {c}^{2} + 2 \times 20 \\ {a}^{2} + {b}^{2} + {c}^{2} = - 39 \\ \\ {a}^{3 } + {b}^{3} + {c}^{3} - 3abc = 1( - 39 - 20) \\ = - 59$