Question

find the value of a^3+b^3+c^3-3ab c when a+b+c=8 and ab+bc+ca= 25​

Answer 1

Step-by-step explanation:

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} -ab - bc - ca) \\ => {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 8{(a + b + c)}^{2} - 2(ab + bc + ca) - (ab + bc + ca) \\ => {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 8{(8)}^{2} - 2(25) - (25) \\ => {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 512 - 50- 25\\ => {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 512 - 75 \\=> {a}^{3} + {b}^{3} + {c}^{3} - 3abc =437$