(a)(i) Factorise \(a^{3}-b^{3}\).
(ii) Factorise \(a^{4}-b^{4}\) as a product of one linear factor and one factor
of degree \(3,\) and as a product of two linear factors and one quadratic
factor. develops the ideas that were implicit in Problem 113. The clue lies in
Problem \(113(\mathrm{a}),\) and in the comment made in the main text in Chapter
1 (after Problem 4 in Chapter 1 ), which we repeat here:
"The last part [of Problem \(113(\mathrm{a})]\) is included to emphasise a
frequently neglected message:
Words and images are part of the way we communicate. But most of us cannot
calculate with words and images.
To make use of mathematics, we must routinely translate words into symbols. So
"numbers" need to be represented by symbols, and points in a geometric diagram
need to be properly labelled before we can begin to calculate, and to reason,
effectively."
As soon as one reads the words "one less than a square", one should
instinctively translate this into the form " \(x^{2}-1\) ". Bells will then
begin to ring; for it is impossible to forget the factorisation
$$
x^{2}-1=(x-1)(x+1)
$$
And it follows that:
for a number that factorises in this way to be prime, the smaller factor \(x-1\)
must be equal to 1 ; \(\therefore x=2,\) so there is only one such prime.
The integer factorisations in Problem \(113(\mathrm{c})-\) namely
$$
3^{3}-1=2 \times 13,4^{3}-1=3 \times 21,5^{3}-1=4 \times 31,6^{3}-1=5 \times
43, \ldots
$$
may help one to remember (or to discover) the related factorisation
$$
x^{3}-1=(x-1)\left(x^{2}+x+1\right)
$$
\(\therefore\) For a number that factorises in this way to be prime, the smaller
factor " \(x-1\) " must be equal to 1 ; \(\therefore x=2,\) so there is only one
such prime.
Problem 113 parts (a) and (c) highlight the completely general factorisation
(Problem \(115(\) a \()(\) iii \()\) :
$$
x^{n}-1=(x-1)\left(x^{n-1}+x^{n-2}+\cdots+x^{2}+x+1\right)
$$
This family of factorisations also shows that we should think about the
factorisation of \(x^{2}-1\) as \((x-1)(x+1),\) with the uniform factor \((x-1)\)
first (rather than as \((x+1)(x-1))\). Similarly, the results of Problem
\(\mathbf{1 1 5}\) show that we should think of the familiar factorisation of
\(a^{2}-b^{2}\) as \((\boldsymbol{a}-\boldsymbol{b})(a+b),(\) not as \((a+b)(a-b),\)
but always with the factor \((\boldsymbol{a}-\boldsymbol{b})\)
first).
The integer factorisations in Problem \(113(\) d \()-\) namely
$$
3^{3}+1=4 \times 7,4^{3}+1=5 \times 13,5^{3}+1=6 \times 21,6^{3}+1=7 \times
31,7^{3}+1=8 \times 43, \ldots
$$
may help one to remember (or to discover) the related factorisation
$$
x^{3}+1=(x+1)\left(x^{2}-x+1\right)
$$
\(\therefore\) For such a number to be prime, one of the factors must be equal
to 1 This time one has to be more careful, because the first bracket may not
be the "smaller factor" - so there are two cases to consider:
(i) if \(x+1=1,\) then \(x=0,\) and \(x^{3}+1=1\) is not prime;
(ii) if \(x^{2}-x+1=1,\) then \(x=0\) or \(x=1,\) so \(x=1\) and we obtain the prime 2
as the only solution.
The factorisation for \(x^{3}+1\) works because " 3 is odd", which allows the
alternating \(+/-\) signs to end in a "+" as required. Hence Problem
\(113(\mathrm{~d})(\mathrm{iii})\) highlights the completely general
factorisation for odd powers:
$$
x^{2 n+1}+1=(x+1)\left(x^{2 n}-x^{2 n-1}+x^{2 n-2}-\cdots+x^{2}-x+1\right)
$$
You probably know that there is no standard factorisation of \(x^{2}+1,\) or of
\(x^{4}+1\) (but see Problem 114 above).
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