Factorize the polynomial a^3 + b ^3 + c^3 - 3abc

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Homework Statement



factorise: a^3 + b^3 + c^3 - 3abc

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The Attempt at a Solution



I had a few random attempts and found that (a + b + c) is factor, and then dividing the original equation by (a + b + c) yeilds a^2 + b^2 + c^2 - ab - ac - bc

I can't figure out how to factorise this quadratic. I tried solving for A and it either doesn;t work, or I did it wrong lol. short of trying a zillion different factors (i'd rather not, I'd prefer a more elegant solution), what can I do?

thanks.
 
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I don't think you can. I stuck it into both maple and mathematica and they return the same factors as you have found.
 
so I've factorised it as far as I can? :S

if so its a stupid problem lol. wouldve thought it would've given linear factors.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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