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How to simplify following fraction? I have tried everything, but nothing seems to work... $$-a^3 (c^2 - b^2) + b^3 (c^2 - a^2) - c^3 (b^2 - a^2)\over (c-b)(c-a)(b-a)$$

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Are there any other relationships between a, b, or c? What is the context of this expression? – Laars Helenius Mar 17 '14 at 21:38
I notice that if $b=c$, then the numerator is zero, so by the factor theorem the numerator is divisible by $b-c.$ Perhaps you can expand the numerator out and use long division to divide the result by $b-c.$ (There's probably a slicker way to do this . . .) – Dave L. Renfro Mar 17 '14 at 21:41
Others are probably going to give slicker methods (but tricker to come up with), so here's what I'm talking about: To divide by $b-c,$ I'll consider $b$ to be the "$x$", so we rewrite the numerator as $(c^2 - a^2)b^3 + (a^3 - c^3)b^2 + (a^2c^3 - a^3c^2)$. The first step in long division is to observe $b$ (the "$x$") divides into $(c^2 - a^2)b^3$ (the leading term in the numerator) $(c^2 - a^2)b^2$ many times. Multiply this by $b-c$ then subtract, then divide $b$ into the new leading term . . . You should be left with a remainder of $0$ after you subtract a second time. – Dave L. Renfro Mar 17 '14 at 21:53
In my rush to get that comment finished before I had to leave yesterday I didn't realize that $3$ subtractions (not $2$) will be needed. I carried out the long division just now and got $(c^2-a^2)b^2 + (a^3-a^2c)b + (a^3c-a^2c^2).$ In the same way that we knew that $b-c$ is a factor of the numerator, we also see that $b-a$ and $a-c$ are factors of the numerator. So now divide the polynomial two sentences back by $b-a,$ and then divide the resulting quotient by $a-c.$ (continued) – Dave L. Renfro Mar 18 '14 at 21:56
(continuation) Of course, this is a bit more tedious than the answers shown, but it has the advantage of being essentially the same method typically encountered in college algebra and precalculus classes. See my answer at Finding limit of a quotient. Finally, it's not really a trick to initially investigate what happens when $b=c,$ $b=a,$ and $a=c,$ since the only factors that could cancel are $b-c,$ $b-a,$ and $a-c$ (nothing else is in the denominator). – Dave L. Renfro Mar 18 '14 at 21:57
up vote 3 down vote accepted

Hands on, the numerator is $$-a^3c^2+a^3b^2+b^3c^2-a^2b^3-c^3b^2+c^3a^2$$Write this as a polynomial in $c$: $$(a^2-b^2)c^3+(b^3-a^3)c^2+a^2b^2(a-b)$$ We can extract a factor of $(a-b)$ to give: $$(a+b)c^3-(a^2+ab+b^2)c^2+a^2b^2=ac^3+bc^3-a^2c^2-abc^2-b^2c^2+a^2b^2$$ Now write this as a polynomial in $b$: $$(a^2-c^2)b^2+(c^3-ac^2)b+ac^2(c-a)$$ from which we extract a factor $(c-a)$ to give:$$-(a+c)b^2+c^2b+ac^2=(c^2-b^2)a+bc(c-b)$$ (written as a polynomial in $a$) from which we take out $(b-c)$ to give $-ab-bc-ca$ So the numerator is $$-(a-b)(b-c)(c-a)(ab+bc+ca)$$

Using the symmetry is the way to go, but note how writing as a polynomial in $c$ highlights the possible factors involving only $a$ and $b$ - which can be handy if you have no better idea how to start.

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Notice that replacing in the numerator $a$ by $b$ or by $c$ we find $0$ and the same thing for $b$ and $c$ so so we can write the numerator on the form $$\lambda(a,b,c)(c-b)(c-a)(b-a)$$ Now to find $\lambda$ let $a=0$ then we have $$b^3c^2-c^3b^2=b^2c^2(b-c)=\lambda(0,b,c)(c-b)cb$$ so $\lambda(0,b,c)=-bc$ and by symmetry we have $$\lambda(a,b,c)=-(ab+bc+cb)$$ and the simplification is straightforward.

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Hint. Start with $$\eqalign{-a^3(c^2-b^2)&+b^3(c^2-a^2)-c^3(b^2-a^2)\cr &=-a^3(c^2-b^2)-(c^3b^2-b^3c^2)+a^2(c^3-b^3)\cr &=-a^3(c+b)(c-b)-b^2c^2(c-b)+a^2(c^2+bc+b^2)(c-b)\ .\cr}$$

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Take advantage of symmetry. First write the expression in the more symmetric form: $$ \frac{a^3(b^2 - c^2) + b^3(c^2 - a^2) + c^3(a^2 - b^2)}{(a-b)(b-c)(c-a)} $$ Notice that

  • The expression is unchanged in any permutation of the variables $(a,b,c)$ (you should check this)

  • Direct substitution of $a = b$ into the numerator gives $0$, meaning the numerator is indeed divisible by the denominator (by symmetry)

  • Every term of the resulting expression must have degree 2, since the numerator is degree 5 and the denominator degree 3.

It follows that the result must be $$ \lambda_1 (a^2 + b^2 + c^2) + \lambda_2 (ab + bc + ca) $$ For some values $\lambda_1$ and $\lambda_2$. Plugging in a simple case like $b = 0$, $c = 1$ you can conclude $\lambda_1 = 0$ and $\lambda_2 = -1$.

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You can also conclude $\lambda_1=0$ and $\lambda_2=-1$ from the fact that, as a rational function in $a$, the symmetric expression is a cubic with lead coefficient $b^2-c^2$ divided by a quadratic with lead coefficient $-(b-c)$. Indeed, you can jump from there straight to the final result, $-(ab+bc+ca)$. – Barry Cipra Mar 17 '14 at 21:52

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