Question

If \[\alpha ,\beta ,\gamma \] are roots of the cubic \[{{x}^{3}}+qx+r=0\], then the equation whose roots are \[{{\left( \alpha -\beta \right)}^{2}},{{\left( \beta -\gamma \right)}^{2}},{{\left( \gamma -\alpha \right)}^{2}}\] is:
a) \[{{y}^{3}}+6q{{y}^{2}}+9{{q}^{2}}y+\left( 4{{q}^{3}}+27{{r}^{2}} \right)=0\]
b) \[{{y}^{3}}+6q{{y}^{2}}+9{{q}^{2}}y+\left( -4{{q}^{3}}-27{{r}^{2}} \right)=0\]
c) \[{{y}^{3}}+6q{{y}^{2}}+9{{q}^{2}}y+\left( 4{{q}^{3}}-27{{r}^{2}} \right)=0\]
d) \[{{y}^{3}}+6q{{y}^{2}}+9{{q}^{2}}y+\left( 4{{r}^{2}}+27{{q}^{3}} \right)=0\]

Answer 1

Hint: In order to find the solution to the given question that is to find the equation whose roots are \[{{\left( \alpha -\beta \right)}^{2}},{{\left( \beta -\gamma \right)}^{2}},{{\left( \gamma -\alpha \right)}^{2}}\] if \[\alpha ,\beta ,\gamma \] are roots of the cubic \[{{x}^{3}}+qx+r=0\], apply the standard formula of to find the cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] from the roots \[\alpha ,\beta ,\gamma \] which satisfy following conditions:\[\alpha \beta \gamma =\dfrac{-d}{a}\]; \[\alpha \beta +\alpha \gamma +\beta \gamma =\dfrac{c}{a}\]; \[\alpha +\beta +\gamma =\dfrac{-b}{a}\].

Complete step by step solution:
According to the question, given cubic equation in the question is as follows:
\[{{x}^{3}}+qx+r=0...\left( 1 \right)\]
We know that \[\alpha ,\beta ,\gamma \] are the roots of the above cubic equation. Now by applying the standard formula of to find the cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] from the roots \[\alpha ,\beta ,\gamma \] which satisfy following conditions:\[\alpha \beta \gamma =\dfrac{-d}{a}\]; \[\alpha \beta +\alpha \gamma +\beta \gamma =\dfrac{c}{a}\]; \[\alpha +\beta +\gamma =\dfrac{-b}{a}\]. Comparing the standard cubic equation with equation \[\left( 1 \right)\] we get:
\[\Rightarrow \alpha \beta \gamma =\dfrac{-d}{a}=-r...\left( 2 \right)\]
\[\Rightarrow \alpha \beta +\alpha \gamma +\beta \gamma =\dfrac{c}{a}=q...\left( 3 \right)\]
\[\Rightarrow \alpha +\beta +\gamma =\dfrac{-b}{a}=0...\left( 4 \right)\]
Now to find the cubic equation from the roots \[{{\left( \alpha -\beta \right)}^{2}},{{\left( \beta -\gamma \right)}^{2}},{{\left( \gamma -\alpha \right)}^{2}}\], we would like to find the three equations:
1) \[{{\left( \alpha -\beta \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}=\dfrac{-B}{A}\];
2) \[{{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}}=\dfrac{C}{A}\] and
3) \[{{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}=\dfrac{-D}{A}\];
So, that the required equation will be of the form: \[{{y}^{3}}+\dfrac{B}{A}{{y}^{2}}+\dfrac{C}{A}y+\dfrac{D}{A}=0...\left( 5 \right)\]

To find these expressions in terms of \[q\] and \[r\], we just have to find the value of the above three equations.
\[{{\left( \alpha -\beta \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta +{{\beta }^{2}}+{{\gamma }^{2}}-2\beta \gamma +{{\alpha }^{2}}+{{\gamma }^{2}}-2\alpha \gamma \]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}=2\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)-2\left( \alpha \beta +\beta \gamma +\alpha \gamma \right)\]
Using equation \[\left( 3 \right)\], we can rewrite that the above equation as:
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}=2\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)-2\left( q \right)\]
We know that \[\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)={{\left( \alpha +\beta +\gamma \right)}^{2}}-2\left( \alpha \beta +\alpha \gamma +\beta \gamma \right),\] therefore by using equation \[\left( 3 \right)\text{ }\!\!\And\!\!\text{ }\left( 4 \right)\], we get \[\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)={{\left( 0 \right)}^{2}}-2\left( q \right)=-2q\].
Now substituting this value in the above equation, we get:
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}=2\left( -2q \right)-2\left( q \right)=-6q\]
\[\Rightarrow \dfrac{B}{A}=-\left( -6q \right)=6q...\left( 6 \right)\]
Now solve for the $2^{nd}$ equation, we get:
\[\begin{align}
  & {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}} \\
 & ={{\left( \left( \alpha -\beta \right)\left( \beta -\gamma \right)+\left( \beta -\gamma \right)\left( \gamma -\alpha \right)+\left( \gamma -\alpha \right)\left( \alpha -\beta \right) \right)}^{2}}-2\left( \left( \alpha -\beta \right)\left( \beta -\gamma \right)\left( \gamma -\alpha \right) \right)\left( \left( \alpha -\beta \right)+\left( \beta -\gamma \right)+\left( \gamma -\alpha \right) \right) \\
\end{align}\]
Clearly, we can see that \[\left( \left( \alpha -\beta \right)+\left( \beta -\gamma \right)+\left( \gamma -\alpha \right) \right)=0\], therefore we can rewrite the above equation as:
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}}={{\left( \left( \alpha -\beta \right)\left( \beta -\gamma \right)+\left( \beta -\gamma \right)\left( \gamma -\alpha \right)+\left( \gamma -\alpha \right)\left( \alpha -\beta \right) \right)}^{2}}\]
Simplify it further by opening the brackets of the terms in the right-hand side of the expression, we get:
\[\Rightarrow {{\left( \alpha \beta -\alpha \gamma -{{\beta }^{2}}+\beta \gamma +\alpha \gamma -{{\alpha }^{2}}-\beta \gamma +\alpha \beta +\beta \gamma -\alpha \beta -{{\gamma }^{2}}+\alpha \gamma \right)}^{2}}\]
Simplify and cancel the terms from the above expression, we get:
\[\Rightarrow {{\left( \left( \alpha \beta +\beta \gamma +\alpha \gamma \right)-\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right) \right)}^{2}}\]
Using equation \[\left( 3 \right)\] and the result \[\left( {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \right)=-2q\] in the above expression we get:
\[\Rightarrow {{\left( \left( q \right)-\left( -2q \right) \right)}^{2}}={{\left( 3q \right)}^{2}}=9{{q}^{2}}\]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}}=9{{q}^{2}}\]
\[\Rightarrow \dfrac{C}{A}=9{{q}^{2}}...\left( 7 \right)\]
At last, solve for the $3^{rd}$ equation,
\[{{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \left( \alpha -\beta \right)\left( \beta -\gamma \right)\left( \gamma -\alpha \right) \right)}^{2}}\]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \left( \alpha -\beta \right)\left( \beta \gamma -\alpha \beta -{{\gamma }^{2}}+\alpha \gamma \right) \right)}^{2}}\]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \alpha \beta \gamma -{{\alpha }^{2}}\beta -\alpha {{\gamma }^{2}}+{{\alpha }^{2}}\gamma -{{\beta }^{2}}\gamma +\alpha {{\beta }^{2}}+\beta {{\gamma }^{2}}-\alpha \beta \gamma \right)}^{2}}\]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \beta {{\gamma }^{2}}-{{\alpha }^{2}}\beta -\alpha {{\gamma }^{2}}+{{\alpha }^{2}}\gamma -{{\beta }^{2}}\gamma +\alpha {{\beta }^{2}} \right)}^{2}}\]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \alpha {{\beta }^{2}}-{{\alpha }^{2}}\beta +{{\alpha }^{2}}\gamma -\alpha {{\gamma }^{2}}+\beta {{\gamma }^{2}}-{{\beta }^{2}}\gamma \right)}^{2}}\]
Take the terms in common within the bracket from right-hand side of the above equation, we get:
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \alpha \beta \left( \alpha -\beta \right)+\alpha \gamma \left( \alpha -\gamma \right)+\beta \gamma \left( \gamma -\beta \right) \right)}^{2}}\]
From equation \[\left( 4 \right)\], we can derive that these result that \[\beta =-\alpha -\gamma \]and \[\gamma =-\alpha -\beta \], now applying these results in the above equation, we get:
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \alpha \beta \left( \alpha -\left( -\alpha -\gamma \right) \right)+\alpha \gamma \left( \alpha -\left( -\alpha -\beta \right) \right)+\beta \gamma \left( \gamma -\left( -\alpha -\gamma \right) \right) \right)}^{2}}\]
Simplify the terms on the right-hand side of the above equation, we get:
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( \alpha \beta \left( 2\alpha +\gamma \right)+\alpha \gamma \left( 2\alpha +\beta \right)+\beta \gamma \left( 2\gamma +\alpha \right) \right)}^{2}}\]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( 2{{\alpha }^{2}}\beta +\alpha \beta \gamma +2{{\alpha }^{2}}\gamma +\alpha \beta \gamma +2\beta {{\gamma }^{2}}+\alpha \beta \gamma \right)}^{2}}\]
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}={{\left( 2{{\alpha }^{2}}\beta +2{{\alpha }^{2}}\gamma +2\beta {{\gamma }^{2}}+3\alpha \beta \gamma \right)}^{2}}\]
Solving the right-hand side of the above equation, we get:
\[\Rightarrow 4\left( \alpha \beta +\beta \gamma +\alpha \gamma \right)\left( \alpha \beta +\beta \gamma +\alpha \gamma \right)\left( \alpha \beta +\beta \gamma +\alpha \gamma \right)+27{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}\]
By applying the result of equation \[\left( 2 \right)\text{ }\!\!\And\!\!\text{ }\left( 3 \right)\], in the above expression, we get:
\[\Rightarrow {{\left( \alpha -\beta \right)}^{2}}{{\left( \beta -\gamma \right)}^{2}}+{{\left( \beta -\gamma \right)}^{2}}{{\left( \gamma -\alpha \right)}^{2}}+{{\left( \gamma -\alpha \right)}^{2}}{{\left( \alpha -\beta \right)}^{2}}=4{{q}^{3}}+27{{r}^{2}}\]
\[\Rightarrow \dfrac{D}{A}=-\left( 4{{q}^{3}}+27{{r}^{2}} \right)=-4{{q}^{3}}-27{{r}^{2}}...\left( 8 \right)\]
Now, substituting the value of equation \[\left( 6 \right),\left( 7 \right)\] and \[\left( 8 \right)\] in the equation \[\left( 5 \right)\], we get:
\[\Rightarrow {{y}^{3}}+6q{{y}^{2}}+9{{q}^{2}}y+\left( -4{{q}^{3}}-27{{r}^{2}} \right)=0\]
Hence, the cubic equation of the roots \[{{\left( \alpha -\beta \right)}^{2}},{{\left( \beta -\gamma \right)}^{2}},{{\left( \gamma -\alpha \right)}^{2}}\] is \[{{y}^{3}}+6q{{y}^{2}}+9{{q}^{2}}y+\left( -4{{q}^{3}}-27{{r}^{2}} \right)=0\]

So, the correct answer is “Option b”.

Note: It’s important to remember that the standard formula of to find the cubic equation \[a{{x}^{3}}+b{{x}^{2}}+cx+d=0\] from the roots \[\alpha ,\beta ,\gamma \] is by satisfying following conditions:\[\alpha \beta \gamma =\dfrac{-d}{a}\]; \[\alpha \beta +\alpha \gamma +\beta \gamma =\dfrac{c}{a}\]; \[\alpha +\beta +\gamma =\dfrac{-b}{a}\].