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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