Can we factorise $a^3(b-c)+b^3(c-a)+c^3(a-b)$?

Can we factorise $a^3(b-c)+b^3(c-a)+c^3(a-b)$?

Question

1. Prove that $\frac{1}{(2-\sqrt{3})^3}+\frac{1}{(2+\sqrt{3})^3}=52.$

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1. Show that $(a-b)$ is a factor of $a^3(b-c)+b^3(c-a)+c^3(a-b)$ and hence factorise the expression completely.

Suggestion

Could we write these two fractions as a single fraction?

What happens if we expand $(a+b)^3 + (a-b)^3?$

Solution

We first write the left hand side as a single fraction, $\frac{1}{(2-\sqrt{3})^3}+\frac{1}{(2+\sqrt{3})^3}=\frac{(2+\sqrt{3})^3 +(2-\sqrt{3})^3}{(2-\sqrt{3})^3(2+\sqrt{3})^3},$ What is the denominator? We have $(2-\sqrt{3})^3(2+\sqrt{3})^3= \bigl((2-\sqrt{3})(2+\sqrt{3})\bigr)^3=(4-3)^3=1^3=1.$

So our fraction becomes $(2+\sqrt{3})^3+(2-\sqrt{3})^3.$

We can use the binomial theorem to write

$$\begin{align*} (a+b)^3 &= a^3 + 3a^2b + 3ab^2 + b^3\\ (a-b)^3 &= a^3 - 3a^2b + 3ab^2 - b^3, \end{align*}$$

so that we have $(a+b)^3+(a-b)^3 = 2a^3 +6ab^2.$ Thus $(2+\sqrt{3})^3 +(2-\sqrt{3})^3 = 2\times 2^3+6\times 2 \times 3 = 52,$ as required.

Let’s define $Q(a,b,c)= a^3(b-c)+b^3(c-a)+c^3(a-b).$

We can show that $(a-b)$ is a factor of this expression using the Factor Theorem.

This says that $(x-p)$ divides the polynomial $P(x)$ if and only if $P(p) = 0$.

So if we regard $Q$ as a polynomial in $a$, then $(a-b)$ will divide $Q(a,b,c)$ if and only if $Q(b,b,c)=0$.

Now $Q(b,b,c)=b^3(b-c)+b^3(c-b)+c^3(b-b)=0$, so $(a-b)$ is a factor of $Q(a,b,c)$.

…and hence factorise the expression completely.

Notice that our expression is such that $Q(a,b,c)= Q(c,a,b) = Q(b,c,a)$ (we say that $Q(a,b,c)$ is symmetrical in $a$, $b$ and $c$).

Because of this symmetry, we know that if $(a-b)$ is a factor of $Q(a,b,c)$, then $(b-c)$ and $(c-a)$ are also.

So $Q(a,b,c)=(a-b)(b-c)(c-a)R(a,b,c)$, where $R$ is a symmetrical polynomial in $a$, $b$ and $c$.

Now $Q(a,b,c)$ is of degree $4$, while $(a-b)(b-c)(c-a)$ is of degree $3$, so $R(a,b,c)$ must be of degree $1$ in $a, b$ and $c$.

The only symmetrical polynomial of degree $1$ is of the form $R(a,b,c)=k(a+b+c)$ where $k$ is a constant. So we have $Q(a,b,c)=(a-b)(b-c)(c-a)k(a+b+c).$

To find the value of $k$, we compare the coefficients of one of the terms. The coefficient of $a^3b$ in $Q(a,b,c)$ is $1$, so $a \times b \times (-a) \times ka = a^3b$, and so $k = -1$.

Thus $Q(a,b,c) = -(a-b)(b-c)(c-a)(a+b+c)$, and this is our factorisation.