Question

please answer this question​

Answer 1

Answer:

$by \: using \: identity \\ {(a + b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca \\ where \\ a + b + c = 10 \\ ab + bc + ca = 8 \\ then \: \\ {a}^{2} + {b}^{2} + {c}^{2} = {(a + b + c)}^{2} - 2(ab + bc + ca) \\ = {10}^{2} - 2 \times 8 \\ 100 - 16 \\ = 84$