+0

# Maths, Algebra, Solution of Equation.

+5
99
2

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

Guest Feb 26, 2017
Sort:

#1
+91256
0

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
+79894
0

(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

### 9 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details