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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.

 Nov 7, 2020
 #1
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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:

 

\(a\in \{0,0,0,0,0\}\\ b\in \{0,1,2,3,9\}\\ c\in \{1,0,9,3,2\}\)

\((a+b+c)\in\{1,1,11,6,11\}\)

laugh  !

 Nov 8, 2020

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