## Miscellaneous Example of Algebra

In order to find values of expressions involving a, b, c when these quantities are connected by the equation $a+b+c=0$, we might employ the substitution:

$a = h + k$

$b = {\omega}h + {\omega^{2}}k$

$c = {\omega^{2}}h + {\omega}k$

If, however, the expressions involve a, b, c symmetrically the method exhibited by the following example is preferable:

Example:

If $a+b+c=0$, show that $6(a^{5}+b^{5}+c^{5}) = 5(a^{3}+b^{3}+c^{3})(a^{2}+b^{2}+c^{2})$

Solution:

We have identically

$(1+ax)(1+bx)(1+cx) = 1 + px + qx^{2}+rx^{3}$, where $p = a+b+c$, $q = ab + bc + ca$, $r = abc$.

Hence, using the condition given, $(1+ax)(1+bx)(1+cx) = 1+qx^{2}+rx^{3}$

Taking logarithms and equating the coefficients of $x^{n}$, we have $\frac{(-1)^{n-1}}{n} (a^{n}+b^{n}+c^{n})$, which is equal to the coefficient of $x^{n}$ in the expansion of $\log {(1+qx^{2}+rx^{3})}$, which is equal to the coefficient of $x^{n}$ in the following:

$(qx^{2} + rx^{3}) - \frac{1}{2}(qx^{2}+rx^{3})^{3} + \frac{1}{3}(qx^{2}+rx^{3})^{3} - \ldots$

By putting $n = 2, 3, 5$ we obtain

$q = -\frac{a^{2} + b^{2} + c^{2}}{2}$

$r = \frac{a^{3} + b^{3} + c^{3}}{3}$

$qr = - \frac{a^{5}+b^{5}+c^{5}}{5}$, and hence, we get

$\frac{a^{5}+b^{5}+c^{5}}{5} = \frac{a^{3}+b^{3}+c^{3}}{3} \times \frac{a^{2}+b^{2}+c^{2}}{2}$, and the required result follows at once.

More stuff to follow…

Nalin Pithwa

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