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# Maths, Algebra, Solution of Equation.

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Please solve this question:- a^3 + b^3 + c^3 -3abc, when ( a+b+c) = 4, & (ab+bc+ca)=7

Guest Feb 26, 2017
#1
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a^3 + b^3 + c^3 -3abc, when ( a+b+c) = 4, & (ab+bc+ca)=7

Is there a minus in front of the a^3 or not???

Try expanding (a+b+c)^3 and see what you get.  I have not done it but I expect it all will become obvious :)

Melody  Feb 26, 2017
#2
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(a + b + c)^3   = 4^3

a^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + b^3 + 3 b^2 c + 3 b c^2 + c^3 =  64

[a^3 + b^3 + c^3] + 3 [ a^2b + a^2c +b^2a + b^2c + c^2a + c^2b] + 6abc  = 64

[a^3 + b^3 + c^3] + 3a^2(b + c) + 3b^2(a + c) + 3c^2(a + b) + 6abc  = 64

[a^3 + b^3 + c^3] + 3a^2(4 - a) + 3b^2(4 - b) + 3c^2(4 - c) + 6abc  = 64

-2[ a^3 + b^3 + c^3] + 12[a^2 + b ^2 + c^2]  + 6abc  =  64     ...... divide by  -2

[ a^3 + b^3 + c^3] - 6[ a^2 + b^2 + c^2] - 3abc  = - 32      (1)

And  (a + b + c)^2  = 16

a^2 + 2 a b + 2 a c + b^2 + 2 b c + c^2  = 16

[a^2 + b^2 + c^2] + 2[ ab + bc + ca]  = 16

[a^2 + b^2 + c^2]  + 2 [ 7 ]  = 16

[a^2 + b^2 + c^2]  + 14  = 16

[a^2 + b^2 + c^2]  = 2     sub  this into  (1)

[ a^3 + b^3 + c^3] - 6[ 2 ] - 3abc  = - 32

[ a^3 + b^3 + c^3] - 12 - 3abc  = - 32

a^3 + b^3 + c^3 - 3abc  =  - 20

CPhill  Feb 26, 2017