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In an exam there is given the general equation for quadratic: $ax^2+bx+c=0$.

It is asking: what does $\dfrac{1}{{x_1}^3}+\dfrac{1}{{x_2}^3}$ equal?

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    what are the numbers $$x_1,x_2$$? – 2017-01-29
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    there are no given numbers – 2017-01-29
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    I think $x_1$ and $x_2$ refer to the roots of the quadratic. – 2017-01-29
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    Do you know Vietas formulas? – 2017-01-29
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    Im stuck with this problem because of the fraction there. I know the vieta's formula but i don't know how to change x1 and x2 to two separate fractions – 2017-01-29
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    the result should be $$\frac{b(3ac-b^2)}{c^3}$$ – 2017-01-29
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    Could you please provide me with some information as to how you solved the equation – 2017-01-29

3 Answers 3

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BIG HINT: $$\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x_1^3+x_2^3}{x_1^3x_2^3}=\frac{(x_1+x_2)(x_1^2-x_1x_2+x_2^2)}{x_1^3x_2^3}=\frac{(x_1+x_2)((x_1+x_2)^2-3x_1x_2)}{(x_1x_2)^3}$$

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    I was thinking where did you get 3X1X2 – 2017-01-29
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    @L.Be Try to expand $(x_1+x_2)^2-3x_1x_2$,and see what you get. – 2017-01-29
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    @kingW3 I tried solving this with sum of cubes and got : abc-b^3/c^3, which is not the correct answer. The correct one is 3abc-b^3/c^3. Still confused with the number 3 there. – 2017-01-29
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    @L.Be Well you have that $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2$ now lets plug this in the equation $x_1^2-x_1x_2+x_2^2=x_1^2+x_2^2-x_1x_2=(x_1+x_2)^2-2x_1x_2-x_1x_2=(x_1+x_2)^2-3x_1x_2$ – 2017-01-29
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Let $\dfrac1{x^3}=y\iff x^3=?$

$$(-c)^3=(ax^2+bx)^3\iff -c^3=a^3(x^3)^2+b^3(x^3)+3ab(x^3)(-c)$$

$$\iff-c^3=\dfrac{a^3}{y^2}+\dfrac{b^3}y-\dfrac{3abc}y$$

$$\iff c^3y^2+(b^3-3abc)y+a^3=0$$ whose roots are $\dfrac1{x_1^3},\dfrac1{x_2^3}$

Can you apply Vieta's formula now?

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If $x_{1}$ and $x_{2}$ are the roots then

$x_{1} + x_{2} = -\frac{b}{a}$ and $x_{1} \cdot x_{2}=\frac{c}{a}$, now $\frac{1}{x_{1}}+\frac{1}{x_{2}} = -\frac{b}{c}$ and $$\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}} = \left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)^3- 3\cdot\frac{1}{x_{1}.x_{2}}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right) = \frac{3abc-b^{3}}{c^{3}}$$