Question

If a, b, c are the roots of ${{x}^{3}}+qx+r=0$, form the equation whose roots are ${{b}^{2}}{{c}^{2}},{{c}^{2}}{{a}^{2}},{{a}^{2}}{{b}^{2}}$.

Answer 1

Hint: A cubic equation of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$, where a, b, c, d are non-zero, has three roots p, q and r. These are related as below,

$p+q+r=\dfrac{-b}{a}$

$pq+qr+rp=\dfrac{c}{a}$

$pqr=-\dfrac{d}{a}$

We can use these relations to get the required equation.

Complete step-by-step answer:

The given cubic equation is ${{x}^{3}}+qx+r=0$. The roots of this equation are given in the question as a, b, c.

We know that in a cubic equation, there is a relation between the roots and the coefficients. It gives the relationship between the sum of roots, products or roots and sum of the product of two roots. Let us consider a general cubic equation of the form $a{{x}^{3}}+b{{x}^{2}}+cx+d=0$, where a, b, c, d are non-zero. We know that a cubic equation has three roots. So let us consider them to be p, q and r. Then, we can write the relationships as below,

$p+q+r=\dfrac{-b}{a}$

$pq+qr+rp=\dfrac{c}{a}$

$pqr=\dfrac{-d}{a}$

Here, the term a is the coefficient of ${{x}^{3}}$, b is the coefficient of ${{x}^{2}}$, c is the coefficient of x and d is the constant.

We have been given the cubic equation as ${{x}^{3}}+qx+r=0$and the roots are given as a, b and c.

So, using the above relationships in the cubic equation given to us, we can write that,

$a+b+c=\dfrac{-(0)}{1}=0.....\left( i \right)$

$abc=\dfrac{-r}{1}=-r....\left( ii \right)$

$ab+bc+ca=\dfrac{q}{1}=q.....\left( iii \right)$

Let us suppose that there is a cubic equation p(x) whose roots are given as ${{a}^{2}}{{b}^{2}},{{b}^{2}}{{c}^{2}}$ and ${{a}^{2}}{{c}^{2}}$.

We also know that if a, b, c are roots of an equation than it can be expressed as,

${{x}^{3}}-\left( a+b+c \right){{x}^{2}}+\left( ab+bc+ca \right)x-abc=0$

So we can write p(x) as, $p(x)={{x}^{3}}-\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}} \right){{x}^{2}}+\left( {{a}^{2}}{{b}^{2}}{{b}^{2}}{{c}^{2}}+{{b}^{2}}{{c}^{2}}{{a}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}{{a}^{2}}{{b}^{2}} \right)x-\left( {{a}^{4}}{{b}^{4}}{{c}^{4}} \right)=0$. Let us name the equation and then we will solve it further.

$p(x)={{x}^{3}}-\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}} \right){{x}^{2}}+\left( {{a}^{2}}{{b}^{4}}{{c}^{2}}+{{b}^{2}}{{c}^{4}}{{a}^{2}}+{{a}^{4}}{{c}^{2}}{{b}^{2}} \right)x-\left( {{a}^{4}}{{b}^{4}}{{c}^{4}} \right)=0....\left( iv \right)$

Now we will use equation (iii) that is given by $ab+bc+ca=q$. We will square both sides, using the result ${{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz$ and we will get,

$\begin{align}

  & {{\left( ab+bc+ca \right)}^{2}}={{q}^{2}} \\

 & {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2a{{b}^{2}}c+2ab{{c}^{2}}+2{{a}^{2}}bc={{q}^{2}} \\

\end{align}$

We will take 2abc common from the last three terms,

${{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2abc(a+b+c)={{q}^{2}}$

Now from equation (i) and (ii) we will substitute the value and get

${{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2r(0)={{q}^{2}}$

${{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}={{q}^{2}}....\left( v \right)$

Now we will use equation (i) given by $a+b+c=0$. Squaring both sides and applying result \[{{\left( x+y+z \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz\], we will get

$\begin{align}

  & {{(a+b+c)}^{2}}=0 \\

 & {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=2ab=2bc+2ac=0 \\

\end{align}$

${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=-2\left( ab+bc+ca \right)$

Now using equation (iii), it will become

${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=-2q.....\left( vi \right)$

Now considering equation (ii) and squaring both sides, we will get

${{\left( abc \right)}^{2}}={{\left( -r \right)}^{2}}$

${{a}^{2}}{{b}^{2}}{{c}^{2}}={{r}^{2}}.....\left( vii \right)$

Again, squaring both sides, we will get

${{a}^{4}}{{b}^{4}}{{c}^{4}}={{r}^{4}}.....\left( viii \right)$

Now we will multiply equation (vi) and equation (vii) as below,

${{a}^{2}}{{b}^{2}}{{c}^{2}}\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)=-2q{{r}^{2}}$

${{a}^{4}}{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{b}^{4}}{{c}^{2}}+{{a}^{2}}{{b}^{2}}{{c}^{4}}=-2q{{r}^{2}}.....\left( ix \right)$

Now, using equation (iv), we can write that

$p(x)={{x}^{3}}-\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}} \right){{x}^{2}}+\left(
{{a}^{2}}{{b}^{4}}{{c}^{2}}+{{b}^{2}}{{c}^{4}}{{a}^{2}}+{{a}^{4}}{{c}^{2}}{{b}^{2}} \right)x-\left( {{a}^{4}}{{b}^{4}}{{c}^{4}} \right)=0$

Substituting the value from equations (v), (viii) and (ix), we will get

$p(x)={{x}^{3}}-{{q}^{2}}{{x}^{2}}-2q{{r}^{2}}x-{{r}^{4}}=0$
Therefore, the equation whose roots are ${{b}^{2}}{{c}^{2}},{{c}^{2}}{{a}^{2}},{{a}^{2}}{{b}^{2}}$ is ${{x}^{3}}-{{q}^{2}}{{x}^{2}}-2q{{r}^{2}}x-{{r}^{4}}=0$.


Note: While using the relationship between zeroes and coefficients of the cubic equation, one must remember that a is the coefficient of ${{x}^{3}}$, b is the coefficient of ${{x}^{2}}$, c is the coefficient of x and d is the constant. So, while writing the relationship between the sum of zeroes of the cubic equation given in the question, i.e. ${{x}^{3}}+qx+r=0$, one must be careful. If the roots are p, q and r, we will get the relation as $p+q+r=\dfrac{-b}{a}$ and here we have b as zero (coefficient of ${{x}^{2}}$). So, one could commit a mistake by considering b = q which could lead to the wrong answer.