I got this problem in a book. While trying to solve this I got something like
$$2(a+b+c)(a^2+b^2+c^2)$$ and can't move forward. Your help will be appreciated.
Asked
Active
Viewed 68 times
0
-
1See this post. – Dietrich Burde May 05 '20 at 16:51
-
Try expanding everything and then reverse-engineering it. – Integrand May 05 '20 at 16:55
-
1Hint: Note that $x+y+z=a+b+c$, $x-y=2(b-a)$, $y-z=2(c-b)$, and $z-x=2(a-c)$. Use Dietrich Burde's link. And note that $$p^2+q^2+r^2-pq-qr-rp=\frac{(p-q)^2+(q-r)^2+(r-p)^2}{2}.$$ – Batominovski May 05 '20 at 16:58
-
1@WETutorialSchool What you wrote as Hint is complete solution – Maverick May 06 '20 at 13:58
-
@Maverick FWIW, I saw a meta post that it is controversial to leave hints as answers. :D – Batominovski May 06 '20 at 14:02
3 Answers
0
$$\sum_{cyc}(x^3-xyz)=(x+y+z)\sum_{cyc}(x^2-xy)=$$ $$=(a+b+c)\sum_{cyc}((a+b-c)^2-(a+b-c)(a+c-b))=$$ $$=(a+b+c)\sum_{cyc}(3a^2-2ab-a^2+b^2+c^2-2bc)=$$ $$=4(a+b+c)\sum_{cyc}(a^2-ab)=4(a^3+b^3+c^3-3abc).$$ I used $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz).$$
Michael Rozenberg
- 194,933
0
$$F=(x^3+y^3+z^3-3xyz)=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)~~~~(1)$$ $$(x+y+z)=(a+b+c)~~~~(2)$$ $$x^2+y^2+z^2=[b^2+c^2+a^2+2bc-2ab-2ac+c^2+a^2+b^2+2ac-2ab-2bc+a^2+b^2+c^2+2ab-2ac-2bc]=[3(a^2+b^2+c^2)-2(ab+bc+ca)]~~~~(3)$$ $$xy+yz+zx=c^2-a^2-b^2+2ab+a^2-b^2-c^2+2bc+c^2-a^2-b^2+2ab$$ $$=[-(a^2+b^2+c^2)+2(ab+bc+ca)]~~~~~(4)$$ So from using (2), (3) and (4) in (1), we get $$F=x^3+y^3+z^3-3xyz=4(a^3+b^3+c^3-3abc).$$
Z Ahmed
- 43,235
0
Note $x+y+z=a+b+c=s$. Then,
$$x^3+y^3+z^3 = (x+y+z)^3-3(x+y)(y+z)(z+x)=s^3-24abc$$
$$3xyz = 3(s-2a)(s-2b)(s-2c)= -3s^3 + 12(ab +bc +ca)s -24abc $$
and
\begin{align} x^3+y^3+z^3 - 3xyz = 4s^3-12(ab +bc +ca)s =4(a^3+b^3+c^3 -3abc) \end{align}
Quanto
- 97,352