Question

if a^2+b^2+c^c = 38 and a+b-c = 6 find (ab+bc-ca)​

Answer 1

Step-by-step explanation:

$a + b - c = 6$

square both sides

$(a + b - c)^{2} = {6}^{2}$

${a}^{2} + {b}^{2} + {c}^{2} + 2ab - 2bc - 2ac = 36$

putting values

$38 + 2(ab - bc - ac) = 36$

$2(ab - bc - ac) = 2$

$(ab - bc - ac) = 1$