5
$\begingroup$

If $a+b+c = 0$ show that $$(2a-b)^3 + (2b-c)^3 + (2c-a)^3 = 3 (2a-b)(2b-c)(2c-a)$$

I have tried substituting the values but it gets too complicated. Can anyone please help with the method? I have been trying for 30 minutes. Thanks!

$\endgroup$
10
$\begingroup$

Let $x=2a-b$, $y=2b-c$, and $z=2c-a$. Then $x+y+z=0$, so $LHS=x^3+y^3+z^3=x^3+y^3+(-x-y)^3=3xy(-x-y)=3xyz=RHS$.

$\endgroup$
  • $\begingroup$ You have explained it beautifully. Thank you so much! $\endgroup$ – Maga Kriv Oct 2 '13 at 2:45
1
$\begingroup$

Hint: $A^3+B^3=(A+B)(A^2-AB+B^2)$, and $(2a-b)+(2b-c)=2a+b-c = a-2c+(a+b+c)=-(2c-a)$

Use these to simplify.

$\endgroup$
0
$\begingroup$

Use Newton's identities for the polynomial with roots $A=2a-b$, $B=2b-c$, $C=2c-a$. Note that $a+b+c = 0$ implies $A+B+C=0$. So, Newton's identities say that $A^3+B^3+C^3=p_3=e_1 p_2 + e_2 p_1 + 3e_3 = 3e_3 = 3ABC$.

Here, $p_k = A^k + B^k + C^k$, $e_1=p_1=A+B+C$, $e_2=AB+BC+CA$ (but is not needed), and $e_3=ABC$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.