It is known that a + B + C = 0, ABC = 8, the square of a + the square of B + the square of C = 32 Connect Find one of a + one of B + one of C We need to be quick

Because (a + B + C) 2 = A2 + B2 + C2 + 2Ab + 2BC + 2Ac
SO 2 (AB + BC + AC) = 0-32
So AB + BC + AC = - 16
So one of a + one of B + one of C = (AB + BC + AC) / ABC = - 16 / 8 = - 2

Given a + B + C = 0, the square of a plus the square of B plus the square of C = 32 ABC = 8, find one part of a plus one part of B plus one part of C

a+b+c=0
(a+b+c)^2=0
a^2+b^2+c^2+2ab+2ac+2bc=0
2ab+2ac+2bc
=(a+b+c)^2-(a^2+b^2+c^2)
=-32
ab+bc+ac=-16
1/a+1/b+1/c
=(ab+ac+bc)/abc
=(-16)/8
=-2

Let a (1,2), B (- 5,8), C (- 2, - 1) be known

AB²=（8﹣2）²+【（﹣5）﹣1】²=72
AC²=【1﹣（﹣2）】²+【2﹣（﹣1）】²=18
BC²=【（﹣5）﹣（﹣2）】²+【8﹣（﹣1）】²=90
So AB & # 178; + AC & # 178; = BC & # 178; (in accordance with Pythagorean theorem)
So ∠ BAC = 90 ° = π / 2

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