How to simpify this?

Question

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

Answer 1

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.

Answer 2

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

Answer 3

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 $\lambda_2 = -1$.

Answer 4

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.