# finding an quadratic equation by the roots & another equation?

I am new to this site & doesn't know any rules & regulations. So sorry if I am doing any mistake.

the question is stated as follows.

I. $\alpha$ and $\beta$ are the roots of the equation $x^2 + bx + c = 0$. find the quadratic equation in terms of $b$ and $c$, whose roots are $\alpha^3$ and $\beta^3$.

II. Hence, find the quadratic equation in terms of $b$ and $c$, whose roots are $\alpha^3 + 1/\beta^3$ and $\beta^3 + 1/\alpha^3$.

I've managed to do the (I) part, but I don't know how to do the (II) part. please tell how to do this.

• You can write down $\alpha$ and $\beta$ in terms of b and c by the well-known formula for the roots of a quadratic equation. Then $(x-\alpha^3)(x-\beta^3)=0$ is an answer to your first question and $(x - \alpha^3 - \frac{1}{\beta^3})(x - \beta^3 - \frac{1}{\alpha^3})=0$ is an answer to your second question. This is only well defined if neither $\alpha$ nor $\beta$ are zero. – user44400 Jul 31 '14 at 14:08
• I appreciate for all of your answers. but in the II part, it says hence. so what shall I do? – Pawan Vimukthi Jul 31 '14 at 14:18
• Please see math notation guide and advice on asking good questions. – user147263 Jul 31 '14 at 15:49

From the given polynomial, yo know the sum $\alpha+\beta=-b$ and product $\alpha\beta=c$ of its roots. You can express the sum $\alpha^3+\beta^3$ and product $\alpha^3\beta^3$ in terms of these: $\alpha^3+\beta^3=(\alpha+\beta)^3-3(\alpha+\beta)\alpha\beta=(-b)^3-3(-b)c$ and $\alpha^3\beta^3=(\alpha\beta)^3=c^3$, which gives you the coefficients for the polynomial in I. Do the same for II, i.e. try to express the sum $\alpha^3+\frac1{\beta^3}+\beta^3+\frac1{\alpha^3}=\alpha^3+\beta^3+\frac{\alpha^3+\beta^3}{\alpha^3\beta^3}$ and product $(\alpha^3+\frac1{\beta^3})(\beta^3+\frac1{\alpha^3})=\alpha^3\beta^3+2+\frac1{\alpha^3\beta^3}$ in terms of $\alpha+\beta$ and $\alpha\beta$; playing around a bit should help.

• hey, how about this, if I say $x = ky\quad y = \alpha^3 + 1/\beta^3, \beta^3 + 1/\alpha^3$, put it in $x^2 + (b^3 - 3bc)x + c^3 = 0$(the final equation in part I), equate to $(y - (\alpha^3 + 1/\beta^3))(y - (\beta^3 + 1/\alpha^3) = 0$, find the k & reput it, can we get the equation required? – Pawan Vimukthi Jul 31 '14 at 15:52

The quadratic equation will have the form:

$$\left (x- \left (\alpha^3+\frac{1}{\beta^3} \right) \right ) \cdot \left (x- \left (\beta^3+\frac{1}{\alpha^3}\right ) \right) \text{ ,with } \alpha, \beta \neq 0$$

You just have to substitute $\alpha$ and $\beta$.

We have $\alpha+\beta=-b$ and $\alpha\beta=c$.

Let $\gamma$ and $\delta$ be $\alpha^3+\frac{1}{\beta^3}$ and $\beta^3+\frac{1}{\alpha^3}$ respectively. If we can find the sum and the product of $\gamma$ and $\delta$ in terms of $b$ and $c$, we will be finished. Note that $$\gamma+\delta=\alpha^3+\beta^3+\frac{\alpha^3+\beta^3}{\alpha^3\beta^3}.$$

Now $\alpha^3+\beta^3=(\alpha+\beta)^3 -3\alpha\beta(\alpha+\beta)$. This can be easily expressed in terms of $b$ and $c$, and therefore so can $\gamma+\delta$.

We leave dealing with $\gamma\delta$ to you.

$(x-\alpha)(x - \beta) = x^2 + bx + c \implies \alpha + \beta = -b,\quad \alpha\beta = c$

If $r$ is a root of the equation $x^2 + bx + c = 0$, Then \begin{align} r^2 + br + c &= 0 \\ r^2 &= -br - c \\ r^3 &= -br^2 - cr \\ &= -b(-br - c) - cr \\ &= b^2r + bc - cr\\ &= (b^2-c)r + bc \end{align}

It follows that \begin{align} \alpha^3 + \beta^3 &= ((b^2-c)\alpha + bc) + ((b^2-c)\beta + bc) \\ &= (b^2-c)(\alpha+\beta) + 2bc \\ &= (b^2-c)(-b) + 2bc \\ &= -b^3 + bc + 2bc \\ &= -b^3 + 3bc \end{align}

and

\begin{align} \alpha^3 \beta^3 &= (\alpha\beta)^3 \\ &= c^3 \end{align}

So the polynomial is $(x-\alpha^3)(x - \beta^3) = x^2 + (b^3 - 3bc)x + c^3$

## Just to check

$(x-2)(x+3) = x^2+x-6$

So $\quad b=1 \quad \text{and}\quad c=-6$

$b^3 - 3bc = 1 + 18 = 19$
$c^3 = -216$

$(x-8)(x+27) = x^2 +19x - 216$