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.
Problem in arithmetic progression
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sequences-and-series
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0Is 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. – 2012-10-06
3 Answers
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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.
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Hint:
We can write $a=b-x$ and $c=b+x$ for some $x$ (Why?)
Now try to use the fomulae
$(s+t)^3 = s^3 + 3s^2t + 3st^2 + t^3$
$(s+t)^2 = s^2 + 2st + t^2$.
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0@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. – 2012-02-22
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Use the fact that $a+c=2b$ to show that
$a^3+4b^3+c^3$
$=(a+c)^3-3ac(a+c)+4b^3$
$=8b^3-6abc+4b^3$
$=3b(4b^2-2ac)$
$=3b((a+c)^2-2ac)$
$=3b(a^2+c^2)$.
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0noted and yes sir! :) – 2012-02-22