Question

315. If at btc and ab + bc + Ca=26,
find a²+b²+c²​

Answer 1

$if \: a + b + c = 26 \: and \: ab + bc + ca = 26 \: then \\ (a + b + c)^{2} = {a}^{2} + b ^{2} + c ^{2} + 2(ab + bc + ca) \\ (26) ^{2} = a ^{2} + b^{2} + c ^{2} + 2(26) \\ 676 = a ^{2} + b^{2} + c ^{2} + 52 \\ 676 - 52 = a ^{2} + b^{2} + c ^{2} \\ 624 = a ^{2} + b ^{2} + c ^{2}$