# Problem in arithmetic progression

Three numbers $a,b,c$ are in arithmetic progression. How do we prove that $a^3 + 4b^3 + c^3 = 3b(a^2 + c^2)$? I need a proof that starts with LHS expression and arrives at RHS expression.

-
Is not this a straight up homework question. From now on, you should try to propose a solution along with your question and not just ask for direct proofs. Cheers. –  Kraken Oct 6 '12 at 10:30

Use the fact that $a+c=2b$ to show that

$a^3+4b^3+c^3$

$=(a+c)^3-3ac(a+c)+4b^3$

$=8b^3-6abc+4b^3$

$=3b(4b^2-2ac)$

$=3b((a+c)^2-2ac)$

$=3b(a^2+c^2)$.

-
thank you Loy!!! –  Thale Feb 22 '12 at 10:31
noted and yes sir! :) –  Thale Feb 22 '12 at 10:35

Hint:

We can write $a=b-x$ and $c=b+x$ for some $x$ (Why?)

Now try to use the fomulae

$(s+t)^3 = s^3 + 3s^2t + 3st^2 + t^3$

$(s+t)^2 = s^2 + 2st + t^2$.

-
i have tried doing the same before posting the question. taking b=(a+c)/2 but couldn't arrive at RHS –  Thale Feb 22 '12 at 8:52
@Pandey: I am not asking you to put $b=(a+c)/2$. I am asking you to replace $a$ and $c$ in terms of $b$ and $x$ and then try. –  Aryabhata Feb 22 '12 at 9:00

Aryabhata is absolutely correct. You should expand on his hint

One more hint $$a = (b-x), c=(b+x)$$

$$a^{3}+4b^{3}+c^{3} = (b-x)^3+4b^3+(b+x)^3$$

$$\Rightarrow (b-x)^{3}+4b^{3}+(b+x)^3 = 6b^{3}+6bx^{2}$$

And expand on that. Anything more will mean giving you the entire answer.

-