Question

If we have an equation \[{{x}^{3}}-1=0\] which has the non-real complex roots a, b then the value of \[{{\left( 1+2\alpha +\beta \right)}^{3}}-{{\left( 3+3\alpha +5\beta \right)}^{3}}\] is?
(A). -7
(B). 6
(C). -5
(D). 0

Answer 1

Hint: Take the equation of cubic x is equal to unity. Now convert it into a known algebraic equation. Now use the general algebraic equation. Now use general algebraic identities to solve this one. Break them into a product of quadratic and linear polynomials. The root of the linear polynomial will be real. But it is given that \[\alpha ,\beta \] are its non – real roots. So, they will be roots of quadratic. Use formula to find roots of quadratic. Now substitute these into the given expression. And solve it by using general algebraic identities, you know beforehand. If \[a{{x}^{2}}+bx+c=0\] is a quadratic, roots of this equation “x” are given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. If a, b are 2 numbers then we have following algebraic identities:
\[\begin{align}
  & {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right) \\
 & {{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right) \\
\end{align}\]

Complete step-by-step solution -
Given equation in the question can be written in terms of x as:
\[\Rightarrow \]\[{{x}^{3}}-1=0\].
By general algebra we have an identity, given as follows:
\[\Rightarrow \]\[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
By applying this on our equation, we get it as:
\[\Rightarrow \left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0\]
By equating first term to zero, we get one root as:
\[\Rightarrow x-1=0\Rightarrow x=1\]
But given \[\alpha ,\beta \] are non – real. So, this is not our required root.
By equating second term to zero, we get other roots as:
\[\Rightarrow {{x}^{2}}+x+1=0\]
For a quadratic equation, given by \[a{{x}^{2}}+bx+c=0\] the roots are given by the formula:
\[\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Here we have a = 1, b = 1, c = 1. By substituting these into formula roots of the equation, \[{{x}^{2}}+x+1=0\] can be written as follows:
\[\Rightarrow x=\dfrac{-1\pm \sqrt{1-4}}{2}=\dfrac{-1\pm \sqrt{3}i}{2}\]
From above, we can say: \[\alpha =\dfrac{-1+\sqrt{3}i}{2}\], \[\beta =\dfrac{-1-\sqrt{3}i}{2}\].
Given expression which we need to solve: - \[{{\left( 1+2\alpha +\beta \right)}^{3}}-{{\left( 3+3\alpha +5\beta \right)}^{3}}\]
By substituting \[\alpha ,\beta \] values, we get the expression:
\[\Rightarrow {{\left( 1+2\left( \dfrac{-1+\sqrt{3}i}{2} \right)+\left( \dfrac{-1-\sqrt{3}i}{2} \right) \right)}^{3}}-{{\left( 3+\dfrac{3\left( -1+\sqrt{3}i \right)}{2}+\dfrac{5\left( -1-\sqrt{3}i \right)}{2} \right)}^{3}}\]
By simplifying the above equation, we get the expression as:
\[\Rightarrow {{\left( 1-1+\sqrt{3}i+\left( \dfrac{-1-\sqrt{3}i}{2} \right) \right)}^{3}}-{{\left( 3-\dfrac{3}{2}-\dfrac{5}{2}+\dfrac{\sqrt{3}i}{2}\left( 3-5 \right) \right)}^{3}}\]
By simplifying the above expression we get the expression as:
\[\Rightarrow {{\left( \dfrac{\sqrt{3}i-1}{2} \right)}^{3}}-{{\left( -1-\sqrt{3}i \right)}^{3}}\]
By simplifying we can write it as:
\[\Rightarrow \dfrac{1}{8}{{\left( \sqrt{3}i-1 \right)}^{3}}+{{\left( \sqrt{3}i+1 \right)}^{3}}\]
By general algebraic techniques, we know the formula as,
\[\begin{align}
  & =\dfrac{1}{8}\left[ 3\sqrt{3}{{i}^{3}}-1-3\sqrt{3}i\left( \sqrt{3i}-1 \right) \right]+\left[ 1+3\sqrt{3}{{i}^{3}}+3\sqrt{3}i\left( 1+\sqrt{3}i \right) \right] \\
 & =\dfrac{1}{8}\left[ -3\sqrt{3}i-1-9{{i}^{2}}+3\sqrt{3}i \right]+\left[ 1-3\sqrt{3}i+3\sqrt{3}i+9{{i}^{2}} \right] \\
\end{align}\]
By simplifying above, we get it as:
\[=\dfrac{1}{8}\left[ -1+9 \right]+\left[ 1-9 \right]\]
By cancelling the common terms, we get the value as:
= 1 – 8 = - 7.
Therefore the value of the given expression is -7.

Note: Be careful, you have to neglect the root x = 1 because it is real and it is given in question the values are non – real. While substituting a, b, c values into the formula be careful don’t forget to write \[\pm \] as we held 2 roots. Generally, students make calculation mistakes in this question as there is a lot to calculate. Do it carefully.