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

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.

- Show that \((a-b)\) is a factor of \[a^3(b-c)+b^3(c-a)+c^3(a-b)\]

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.