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