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Question Number 12126 by tawa last updated on 13/Apr/17

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 16/Apr/17

a^2 +b^2 +c^2 +2ab+2ac+2bc=0 ⇒ab+bc+ca=−21 A=(a−b)(b−c)(c−a) (a+b+c)^3 =(a+b)^3 +3(a+b+c)(a+b)c+c^3 =a^3 +3ab(a+b)+b^3 +c^3 =105+3ab(−c)=0 ⇒abc=35 x^3 −(a+b+c)x^2 +(ab+bc+ca)x−abc=0 x^3 −21x−35=0 a=5.25,b=−3.13,c=−2.12 a−b=8.38,b−c=−1.01,c−a=−7.37 A=(8.38)(−1.01)(−7.37)=62.37#63.

a2+b2+c2+2ab+2ac+2bc=0⇒ab+bc+ca=−21A=(a−b)(b−c)(c−a)(a+b+c)3=(a+b)3+3(a+b+c)(a+b)c+c3=a3+3ab(a+b)+b3+c3=105+3ab(−c)=0⇒abc=35x3−(a+b+c)x2+(ab+bc+ca)x−abc=0x3−21x−35=0a=5.25,b=−3.13,c=−2.12a−b=8.38,b−c=−1.01,c−a=−7.37You can't use 'macro parameter character #' in math mode

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