Answer:
Step-by-step explanation:
a+b+c=0, a+b=-c,a+c=-b, b+c=-a
(a+b+c)^3=(a+b+c)^2*(a+b+c)=(a^2+b^2+c^2+2ab+2ac+2bc)*(a+b+c)=
a^3+ab^2+ac^2+2a^2b+2a^2c+2abc+a^2b+b^3+bc^2+2ab^2+2abc+2b^2c+a^2c+b^2c+c^3+2abc+2ac^2+2bc^2=a^3+b^3+c^3+3a^2b+3a^2c+3ac^2+3ab^2+3bc^2+3b^2c+6abc=
a^3+b^3+c^3+3a^2*(b+c)+3c^2(a+b)+3b^2(a+c)+6abc=
a^3+b^3+c^3+3a^2*(-a)+3c^2*(-c)+3b^2*(-b)+6abc=
a^3+b^3+c^3-3a^3-3c^3-3b^3+6abc=
6abc-2a^3-2b^3-2c^3=2(3abc-a^3-b^3-c^3)=
2*[3abc-(a^3+b^3+c^3)]=0
so 3abc-(a^3+b^3+c^3)=0
so a^3+b^3+c^3=3abc
I know the 1st and last one but i’m not sure on the rest sorry :(
Answer:
invite taka kung and I have to go out for the night as well and
Step-by-step explanation:
sige and sa couple member a said weeks the a of days is will
A way to add fractions that always works is to multiply each numerator by the denominator of the other, then express the sum of products over the product of the denominators.
Here, you have
The sum is -1 1/12
