Question

If a+b+c = 5 and ab+bc+can = 10, then prove that a^3 + b^3 + c^3 - 3xyz = -25

Answer 1

Answer:

✅Here's your answer

Given

a+b+c= 5

ab +bc+ca =10

By using the identity.....

${a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c) \\ ( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ \\ - 25 = (5)( {a}^{2} + {b}^{2} + {c}^{2} - (10)) \\ \\ - 25 = 5( {a}^{2} + {b}^{2} + {c}^{2} - 10) \\ \\ - 25 = 5 {a}^{2} + 5 {b}^{2} + 5 {c}^{2} - 50 \\ - 25 + 50 = 5( {a}^{2} + {b}^{2} + {c}^{2})$

$25 = 5( {a}^{2} + {b}^{2} + {c}^{2} ) \\ \\ 5 = {a}^{2} + {b}^{2} + {c}^{2}$

✔by using this value u can also prove the question......

Hope it helps ✌✌...