If $\alpha \,and\,\beta $ are the roots of quadratic equation $a{{x}^{2}}+bx+c=0$ then find the value of \[{{\left( a\alpha +b \right)}^{-3}}+{{\left( a\beta +b \right)}^{-3}}\].
Answer
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Hint: Firstly write the values of sum and product of the roots and then simplify the equation by using the formula \[{{a}^{-n}}=\dfrac{1}{{{a}^{n}}}\] and take ‘a’ common. Then use the values of sum and product of roots and then simplify the equation till you get \[S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( -\beta \right)}^{3}}}+\dfrac{1}{{{\left( -\alpha \right)}^{3}}} \right]\]. Then simplify the equation and used the formula \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\] and the values of sum and products of roots to get the final answer.
Complete step by step answer:
To solve the given question we will write down the given data first, therefore,
$a{{x}^{2}}+bx+c=0$ and $\alpha \,and\,\beta $ and the roots of the given quadratic equation.
TO proceed further in the solution we should know the formula given below,
Formula:
If \[p{{x}^{2}}+qx+r=0\] is a quadratic equation having its roots as ‘m’ and ‘n’ then,
\[m+n=\dfrac{-q}{p}\] and \[mn=\dfrac{r}{p}\]
By using the formula given above we can write the sum and products of roots of quadratic equation $a{{x}^{2}}+bx+c=0$ as follows,
\[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\] ……………………………………… (1)
Now we will write the expression given in the question and assume it as ‘S’,
\[\therefore S={{\left( a\alpha +b \right)}^{-3}}+{{\left( a\beta +b \right)}^{-3}}\] ……………………………………………………………. (2)
As we know that \[{{a}^{-n}}=\dfrac{1}{{{a}^{n}}}\] therefore by using this formula in the above equation we will get,
\[\therefore S=\dfrac{1}{{{\left( a\alpha +b \right)}^{3}}}+\dfrac{1}{{{\left( a\beta +b \right)}^{3}}}\]
If we take ‘a’ common from the above equation we will get,
\[\therefore S=\dfrac{1}{{{\left[ a\left( \alpha +\dfrac{b}{a} \right) \right]}^{3}}}+\dfrac{1}{{{\left[ a\left( \beta +\dfrac{b}{a} \right) \right]}^{3}}}\]
Above equation can also be simplified as,
\[\therefore S=\dfrac{1}{{{a}^{3}}{{\left( \alpha +\dfrac{b}{a} \right)}^{3}}}+\dfrac{1}{{{a}^{3}}{{\left( \beta +\dfrac{b}{a} \right)}^{3}}}\]
If we observe the above equation carefully then we will see that all the terms of above equation has \[\dfrac{1}{{{a}^{3}}}\] and therefore we can take it out therefore we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha +\dfrac{b}{a} \right)}^{3}}}+\dfrac{1}{{{\left( \beta +\dfrac{b}{a} \right)}^{3}}} \right]\]
We can replace \[\dfrac{b}{a}\] by \[-\left( -\dfrac{b}{a} \right)\] as both have same values therefore by replacing in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha -\left( -\dfrac{b}{a} \right) \right)}^{3}}}+\dfrac{1}{{{\left( \beta -\left( -\dfrac{b}{a} \right) \right)}^{3}}} \right]\]
If we put the values of equation (1) in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha -\left( \alpha +\beta \right) \right)}^{3}}}+\dfrac{1}{{{\left( \beta -\left( \alpha +\beta \right) \right)}^{3}}} \right]\]
Bu opening the brackets in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha -\alpha -\beta \right)}^{3}}}+\dfrac{1}{{{\left( \beta -\alpha -\beta \right)}^{3}}} \right]\]
By simplifying the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( -\beta \right)}^{3}}}+\dfrac{1}{{{\left( -\alpha \right)}^{3}}} \right]\]
Further simplification in the above equation will give,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-1}{{{\beta }^{3}}}+\dfrac{-1}{{{\alpha }^{3}}} \right]\]
By performing addition in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-{{\alpha }^{3}}-{{\beta }^{3}}}{{{\beta }^{3}}{{\alpha }^{3}}} \right]\]
Further simplification in the above equation will give,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)}{{{\left( \beta \alpha \right)}^{3}}} \right]\]
Now to proceed further in the solution we should know the formula given below,
Formula:
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\]
Above formula can also be written as,
\[{{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)\]
By using above formula in ‘S’ we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-\left( {{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right) \right)}{{{\left( \beta \alpha \right)}^{3}}} \right]\]
Above equation can also be written as,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{{{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right)}{{{\left( \alpha \beta \right)}^{3}}} \right]\]
If we put the values of equation (1) in the above equation we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{{{\left( \dfrac{-b}{a} \right)}^{3}}-3\times \dfrac{c}{a}\times \left( \dfrac{-b}{a} \right)}{{{\left( \dfrac{c}{a} \right)}^{3}}} \right]\]
Further simplification in the above equation we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}-\left( \dfrac{-3bc}{{{a}^{2}}} \right)}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
If we multiply both the numerator and denominator of \[\dfrac{-3bc}{{{a}^{2}}}\] by ‘a’ we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}-\left( \dfrac{-3bc}{{{a}^{2}}} \right)\times \dfrac{a}{a}}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}-\left( \dfrac{-3abc}{{{a}^{3}}} \right)}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
By opening the bracket in the above equation we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}+\dfrac{3abc}{{{a}^{3}}}}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}+3abc}{{{a}^{3}}}}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{-{{b}^{3}}+3abc}{{{a}^{3}}}\times \dfrac{{{a}^{3}}}{{{c}^{3}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ -{{b}^{3}}+3abc\times \dfrac{1}{{{c}^{3}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{-{{b}^{3}}+3abc}{{{c}^{3}}} \right]\]
If we open the brackets we will get,
\[\therefore S=\dfrac{-1\left( -{{b}^{3}} \right)+\left( -1 \right)\left( 3abc \right)}{{{a}^{3}}{{c}^{3}}}\]
\[\therefore S=\dfrac{{{b}^{3}}-3abc}{{{a}^{3}}{{c}^{3}}}\]
If we compare the above equation with equation (2) we will get,
\[\therefore {{\left( a\alpha +b \right)}^{-3}}+{{\left( a\beta +b \right)}^{-3}}=\dfrac{{{b}^{3}}-3abc}{{{a}^{3}}{{c}^{3}}}\]
Therefore the value of \[{{\left( a\alpha +b \right)}^{-3}}+{{\left( a\beta +b \right)}^{-3}}\] is equal to \[\dfrac{{{b}^{3}}-3abc}{{{a}^{3}}{{c}^{3}}}\].
Note: Don’t use the formula \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\] as if you use it then still you will lengthen your answer and still you will get confused. You can use the formula \[{{a}^{3}}+{{b}^{3}}=={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)\] as I have derived it in the above example.
Complete step by step answer:
To solve the given question we will write down the given data first, therefore,
$a{{x}^{2}}+bx+c=0$ and $\alpha \,and\,\beta $ and the roots of the given quadratic equation.
TO proceed further in the solution we should know the formula given below,
Formula:
If \[p{{x}^{2}}+qx+r=0\] is a quadratic equation having its roots as ‘m’ and ‘n’ then,
\[m+n=\dfrac{-q}{p}\] and \[mn=\dfrac{r}{p}\]
By using the formula given above we can write the sum and products of roots of quadratic equation $a{{x}^{2}}+bx+c=0$ as follows,
\[\alpha +\beta =\dfrac{-b}{a}\] and \[\alpha \beta =\dfrac{c}{a}\] ……………………………………… (1)
Now we will write the expression given in the question and assume it as ‘S’,
\[\therefore S={{\left( a\alpha +b \right)}^{-3}}+{{\left( a\beta +b \right)}^{-3}}\] ……………………………………………………………. (2)
As we know that \[{{a}^{-n}}=\dfrac{1}{{{a}^{n}}}\] therefore by using this formula in the above equation we will get,
\[\therefore S=\dfrac{1}{{{\left( a\alpha +b \right)}^{3}}}+\dfrac{1}{{{\left( a\beta +b \right)}^{3}}}\]
If we take ‘a’ common from the above equation we will get,
\[\therefore S=\dfrac{1}{{{\left[ a\left( \alpha +\dfrac{b}{a} \right) \right]}^{3}}}+\dfrac{1}{{{\left[ a\left( \beta +\dfrac{b}{a} \right) \right]}^{3}}}\]
Above equation can also be simplified as,
\[\therefore S=\dfrac{1}{{{a}^{3}}{{\left( \alpha +\dfrac{b}{a} \right)}^{3}}}+\dfrac{1}{{{a}^{3}}{{\left( \beta +\dfrac{b}{a} \right)}^{3}}}\]
If we observe the above equation carefully then we will see that all the terms of above equation has \[\dfrac{1}{{{a}^{3}}}\] and therefore we can take it out therefore we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha +\dfrac{b}{a} \right)}^{3}}}+\dfrac{1}{{{\left( \beta +\dfrac{b}{a} \right)}^{3}}} \right]\]
We can replace \[\dfrac{b}{a}\] by \[-\left( -\dfrac{b}{a} \right)\] as both have same values therefore by replacing in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha -\left( -\dfrac{b}{a} \right) \right)}^{3}}}+\dfrac{1}{{{\left( \beta -\left( -\dfrac{b}{a} \right) \right)}^{3}}} \right]\]
If we put the values of equation (1) in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha -\left( \alpha +\beta \right) \right)}^{3}}}+\dfrac{1}{{{\left( \beta -\left( \alpha +\beta \right) \right)}^{3}}} \right]\]
Bu opening the brackets in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( \alpha -\alpha -\beta \right)}^{3}}}+\dfrac{1}{{{\left( \beta -\alpha -\beta \right)}^{3}}} \right]\]
By simplifying the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{1}{{{\left( -\beta \right)}^{3}}}+\dfrac{1}{{{\left( -\alpha \right)}^{3}}} \right]\]
Further simplification in the above equation will give,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-1}{{{\beta }^{3}}}+\dfrac{-1}{{{\alpha }^{3}}} \right]\]
By performing addition in the above equation we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-{{\alpha }^{3}}-{{\beta }^{3}}}{{{\beta }^{3}}{{\alpha }^{3}}} \right]\]
Further simplification in the above equation will give,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-\left( {{\alpha }^{3}}+{{\beta }^{3}} \right)}{{{\left( \beta \alpha \right)}^{3}}} \right]\]
Now to proceed further in the solution we should know the formula given below,
Formula:
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\]
Above formula can also be written as,
\[{{a}^{3}}+{{b}^{3}}={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)\]
By using above formula in ‘S’ we will get,
\[\therefore S=\dfrac{1}{{{a}^{3}}}\left[ \dfrac{-\left( {{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right) \right)}{{{\left( \beta \alpha \right)}^{3}}} \right]\]
Above equation can also be written as,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{{{\left( \alpha +\beta \right)}^{3}}-3\alpha \beta \left( \alpha +\beta \right)}{{{\left( \alpha \beta \right)}^{3}}} \right]\]
If we put the values of equation (1) in the above equation we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{{{\left( \dfrac{-b}{a} \right)}^{3}}-3\times \dfrac{c}{a}\times \left( \dfrac{-b}{a} \right)}{{{\left( \dfrac{c}{a} \right)}^{3}}} \right]\]
Further simplification in the above equation we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}-\left( \dfrac{-3bc}{{{a}^{2}}} \right)}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
If we multiply both the numerator and denominator of \[\dfrac{-3bc}{{{a}^{2}}}\] by ‘a’ we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}-\left( \dfrac{-3bc}{{{a}^{2}}} \right)\times \dfrac{a}{a}}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}-\left( \dfrac{-3abc}{{{a}^{3}}} \right)}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
By opening the bracket in the above equation we will get,
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}}{{{a}^{3}}}+\dfrac{3abc}{{{a}^{3}}}}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{\dfrac{-{{b}^{3}}+3abc}{{{a}^{3}}}}{\dfrac{{{c}^{3}}}{{{a}^{3}}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{-{{b}^{3}}+3abc}{{{a}^{3}}}\times \dfrac{{{a}^{3}}}{{{c}^{3}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ -{{b}^{3}}+3abc\times \dfrac{1}{{{c}^{3}}} \right]\]
\[\therefore S=\dfrac{-1}{{{a}^{3}}}\left[ \dfrac{-{{b}^{3}}+3abc}{{{c}^{3}}} \right]\]
If we open the brackets we will get,
\[\therefore S=\dfrac{-1\left( -{{b}^{3}} \right)+\left( -1 \right)\left( 3abc \right)}{{{a}^{3}}{{c}^{3}}}\]
\[\therefore S=\dfrac{{{b}^{3}}-3abc}{{{a}^{3}}{{c}^{3}}}\]
If we compare the above equation with equation (2) we will get,
\[\therefore {{\left( a\alpha +b \right)}^{-3}}+{{\left( a\beta +b \right)}^{-3}}=\dfrac{{{b}^{3}}-3abc}{{{a}^{3}}{{c}^{3}}}\]
Therefore the value of \[{{\left( a\alpha +b \right)}^{-3}}+{{\left( a\beta +b \right)}^{-3}}\] is equal to \[\dfrac{{{b}^{3}}-3abc}{{{a}^{3}}{{c}^{3}}}\].
Note: Don’t use the formula \[{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)\] as if you use it then still you will lengthen your answer and still you will get confused. You can use the formula \[{{a}^{3}}+{{b}^{3}}=={{\left( a+b \right)}^{3}}-3ab\left( a+b \right)\] as I have derived it in the above example.
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