Maths Extension Question (1 Viewer)

181jsmith · Nov 7, 2011

The polynomial equation $x^{3}+bx^{2}+cx+d=0$ has roots $a,a^{2},a^{3}$ for some real number a $\neq$ 0.

i) Find in terms of b,c and d the value of $\frac{1}{a}+\frac{1}{a^{2}}+\frac{1}{a^{3}}$

ii) Show that $b^{3}d-c^{3}=0$

Skeptyks · Nov 7, 2011

Sorry, i didn't notice it was a power not a sub..whatever its called.

Alkanes · Nov 8, 2011

Ok so

i)Roots of $x^3+bx^2+cx+d=0$ are $a, a^2, a^3$

Hence -b = $a+a^2+a^3$ , c = $a^3+a^4+a^5$ and -d= $a^6$

Therefore $\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=\frac{a^5+a^4+a^3}{a^6}=\frac{-c}{d}$

ii) c = $-a^2b$

Therefore $c^3$ = $-a^6b^3$

Hence $c^3$ = $db^3$

Therefore, $b^3d-c^3$ = 0

SpiralFlex · Nov 8, 2011

I am sure I answered this question a while back for you...? Or was that for someone else...?

Maths Extension Question (1 Viewer)

181jsmith

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Skeptyks

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Alkanes

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SpiralFlex

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