Question

If 1,$\alpha$,${\alpha }^2$,${\alpha }^3$,............,${\alpha }^{26}$ are the roots of the equation x = $(1{)}^{\frac{1}{27}}$.Find the product of the roots

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Answer 1

If 1,$\alpha$,${\alpha }^2$,${\alpha }^3$,............,${\alpha }^{26}$ are roots of the equation x = $(1{)}^{\frac{1}{27}}$

Given equation $x^{27}$ - 1 = 0

Product of the roots =1 $\times$ $\alpha$ $\times$ ${\alpha }^2$ $\times$ ${\alpha }^3$ $\times$ ............,${\alpha }^{26}$

= ${\alpha }^{(1+2+3+.....26)}$

= $\alpha \frac{26\times 27}{2}$ = ${\alpha }^{351}$

= $(cos\frac{2\pi }{27}+isin\frac{2\pi }{27}{)}^{351}$

= $cos\frac{2\pi \times 351}{27}+isin\frac{2\pi }{27}\times 351$

= cos26$\pi$ + isin26$\pi$

= 1

Alternative

$\alpha$ = $(cos\frac{2\pi k}{n}+isin\frac{2k\pi }{n})$

k = 0,1,2,3,4........

1,$\alpha$,${\alpha }^2$,${\alpha }^3$,${\alpha }^4$............,${\alpha }^{26}$

= $(-1{)}^{n-1}$ = $(-1{)}^{27-1}$ = $(-1{)}^{26}$ = 1