# Algebra

For many people, especially as their schooldays recede further into the past, multiplication or division (or Uglification and Derision, as it has been called) of one *number* by another is tricky enough without a calculator.

But multiplication or division of one *algebraic expression* by another might be regarded as impossible or even meaningless. In fact they're not really very different at all from their numerical counterparts.

To prove that (a^{2} – b^{2}) = (a – b)*(a + b) we evaluate the right-hand side by multiplying each element in the second bracket by each element in the first bracket, and then adding-up the results, remembering that 'a times b' is the same as 'b times a'.

- (a – b)*(a + b) = a*a + a*b – b*a – b*b = a*a - b*b = a
^{2}– b^{2}

Well that's all the homework required.

But one might well wonder if something similar holds for (a^{3} – b^{3}) or (a^{4} – b^{4}), and so on. And indeed it does. But we've not been given the answer in advance – in other words, we have to discover something rather than simply prove it.

In fact, we can work out (a^{4} – b^{4}) straight away from what we've already proved.

- (a
^{4}– b^{4}) = ( (a^{2})^{2}– (b^{2})^{2}) = (a^{2}– b^{2}) * (a^{2}+ b^{2}) = (a – b) * (a + b) * (a^{2}+ b^{2})

Indeed, we could take this a step further by introducing j which is defined as the square-root of -1.

- j = √(-1) and so j
^{2}= -1

- (a – jb)*(a + jb) = a*a + a*jb – jb*a – j
^{2}b*b = a*a - j^{2}b*b = a^{2}+ b^{2}

and so

- (a
^{4}– b^{4}) = (a – b) * (a + b) * (a^{2}+ b^{2}) = (a – b) * (a + b) *( a – jb) * (a + jb)

But this is a bit of a digression.

Cutting to the chase, we haven't yet tackled the prior case of (a^{3} – b^{3}). However we feel pretty confident that it will involve (a – b). As it does.

To discover the other ingredients, we must divide (a^{3} – b^{3}) by (a – b) in exactly the same way as a numerical long-division, dividing the 'a' of a – b firstly into the a^{3} then into the a^{2}b and finally into the ab^{2}, at which point the process terminates. The answer accumulates over on the right-hand side:

a – b ) | a^{3} – b^{3} | ( a^{2} + a*b + b^{2} |

a^{3} – a^{2} b | ||

a^{2} b – b^{3} | ||

a^{2} b – ab^{2} | ||

ab^{2} – b^{3} | ||

ab^{2} – b^{3} | ||

and so

- (a
^{3}– b^{3}) = (a – b) * ( a^{2}+ a*b + b^{2})

Exactly analogous long-divisions are possible for all higher odd powers of 'a' and 'b'

- (a
^{5}– b^{5}) = (a – b) * ( a^{4}+ a^{3}b + a^{2}b^{2}+ ab^{3}+ b^{4})

- (a
^{7}– b^{7}) = (a – b) * ( a^{6}+ a^{5}b + a^{4}b^{2}+ a^{3}b^{3}+ a^{2}b^{4}+ ab^{5}+ b^{6})

- etc

And indeed for all higher even powers of 'a' and 'b' too

- (a
^{6}– b^{6}) = (a – b) * ( a^{5}+ a^{4}b + a^{3}b^{2}+ a^{2}b^{3}+ ab^{4}+ b^{5})

- (a
^{8}– b^{8}) = (a – b) * ( a^{7}+ a^{6}b + a^{5}b^{2}+ a^{4}b^{3}+ a^{3}b^{4}+ a^{2}b^{5}+ ab^{6}+ b^{7})

- etc

but we can use the same simplifications as before

- (a
^{6}– b^{6}) = ((a^{3})^{2}– (b^{3})^{2}) = (a^{3}– b^{3}) * (a^{3}+ b^{3}) = (a – b) * ( a^{2}+ a*b + b^{2}) * (a^{3}+ b^{3})

- (a
^{8}– b^{8}) = ((a^{4})^{2}– (b^{4})^{2}) = (a^{4}– b^{4}) * (a^{4}+ b^{4}) = (a^{2}– b^{2}) * (a^{2}+ b^{2}) * (a^{4}+ b^{4}) = (a – b ) * (a + b) * (a^{2}+ b^{2}) * (a^{4}+ b^{4})

(Eboracus just *loves* type-setting all this sort of stuff into html!)