Given that 2(-a-b-c)^3-27/8=9(-a-b-c)(ab+bc+ca) and 2(a+b+c)-4(ab+bc+ca)+6abc=-9(a+b+c)+9(ab+bc+ca)-9abc+9, find a+b+c.
+14903 Given that
\(2(-a-b-c)^3-27/8=9(-a-b-c)(ab+bc+ca\)
and
\(4(ab+bc+ca)+6abc=-9(a+b+c)+9(ab+bc+ca)-9abc+9 \)
Find a+b+c.
Hello Guest!
\( 2(-a-b-c)^3-27/8=9(-a-b-c)(ab+bc+ca)\)
wolframalpha.com:
\(a\approx -1.11502\\ b\approx 0.162679\\ c\approx -1.75387\\ \color{blue}a+b+c\approx -2.706211\)
\(4(ab+bc+ca)+6abc=-9(a+b+c)+9(ab+bc+ca)-9abc+9 \)
wolframalpha.com
Integer solutions:
\(a\in \{0,0,0,0,0\}\\ b\in \{0,1,2,3,9\}\\ c\in \{1,0,9,3,2\}\)
\((a+b+c)\in\{1,1,11,6,11\}\)
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