Question

If $f(x) = 27{x^3} + \dfrac{1}{{{x^3}}}$ and \[\alpha \],$\beta $ are the roots of$3x + \dfrac{1}{x} = 2$
A). $f(\alpha ) = f(\beta )$
b). $f(\alpha )=-10$
C). $f(\beta )=-10$
D). None of these

Answer 1

Hint: First we have to define what the terms we need to solve the problem are. \[\alpha \],$\beta $are the roots given equation $3x + \dfrac{1}{x} = 2$
Since ${(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$
Here we are give the roots of linear equation which means the value of x will be \[\alpha \] & $\beta $ so we put the value of x in the polynomial as we have the polynomial in the same variable x.

Complete step-by-step solution:
It is given that $f(x) = 27{x^3} + \dfrac{1}{{{x^3}}}$ be the functions of the polynomial .
And \[\alpha \],$\beta $ are the roots of $3x + \dfrac{1}{x} = 2$
Let us take cubic roots on both sides we get,
${(3x+\dfrac{1}{x})^3}={2^3}$ ……………………….. $(1)$
Since as we know the formula for ${(a + b)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$
By applying this formula in equation $(1)$ right hand side, we get
${(3x + \dfrac{1}{x})^3} = 27{x^3} + \dfrac{1}{{{x^3}}} + 3{(3x)^2}(\dfrac{1}{x}) + 3(3x){(\dfrac{1}{x})^2}$
Solving above equations by cancelling $x$ and $\dfrac{1}{x}$ so that we can simplify this equation Only on the left hand side
${(3x + \dfrac{1}{x})^3} = 27{x^3} + \dfrac{1}{{{x^3}}} + 3(3x)(\dfrac{1}{x})(3x + \dfrac{1}{x})$
$\Rightarrow 27{x^3} + \dfrac{1}{{{x^3}}} + 9(3x + \dfrac{1}{x})$
$\Rightarrow 27{x^3} + \dfrac{1}{{{x^3}}} + 9(2)$ [Since from given that$3x + \dfrac{1}{x} = 2$]
Substitute this equation only on the left hand side in $(1)$, we get
${(3x + \dfrac{1}{x})^3} = 27{x^3} + \dfrac{1}{{{x^3}}} + 9(2)$ And in right hand we have ${2^3} = 8$
Combining these two equations from left and right hand sides, we get
 $27{x^3} + \dfrac{1}{{{x^3}}} + 9(2) = 8$
$ \Rightarrow 27{x^3} + \dfrac{1}{{{x^3}}} + 18 = 8$($18$ will go on the right hand side so that we can cancel $8 - 18$)
$ \Rightarrow 27{x^3} + \dfrac{1}{{{x^3}}} = - 10$ which is the polynomial function of the given equations $(2)$
Since \[\alpha \],$\beta $are the roots given equation $3x + \dfrac{1}{x} = 2$
Now we are going to find the roots of the given cubic polynomial so
Put polynomial $x = \alpha $in equation $(2)$we get $ \Rightarrow 27{\alpha ^3} + \dfrac{1}{{{\alpha ^3}}} = - 10$ $(3)$
Is the equation of the cubic root with respect to \[\alpha \]
Put polynomial $x = \beta $in equation $(2)$ we get $ \Rightarrow 27{\beta ^3} + \dfrac{1}{{{\beta ^3}}} = - 10$ $(4)$
Is the equation of the cubic root with respect to $\beta $
Thus from given polynomial $f(x) = 27{x^3} + \dfrac{1}{{{x^3}}}$and put $x = \alpha $ then we get
$f(\alpha ) = 27{\alpha ^3} + \dfrac{1}{{{\alpha ^3}}}$ apply equation $(3)$ in here we get$f(\alpha ) = - 10$
Similarly from given polynomial $f(x) = 27{x^3} + \dfrac{1}{{{x^3}}}$and put $x = \beta $ then we get
$f(\alpha ) = 27{\beta ^3} + \dfrac{1}{{{\beta ^3}}}$apply equation $(4)$ in here we get$f(\beta ) = - 10$
Hence if you see $f(\alpha ) = - 10$ and $f(\beta ) = - 10$
Also $f(\alpha ) = f(\beta )$
Hene Option A, B and C are correct options.

Note: $f(x)$ is the polynomial function as the polynomials have degree 3 so the cubic polynomial and \[\alpha \], $\beta $ are the roots of the linear equation. So for simplifying we use the algebraic identities as:
$(a + b)(a + b)(a + b) = {(a + b)^2}(a + b) = {(a + b)^3}$
\[\alpha \], $\beta $ are the roots given equation $3x + \dfrac{1}{x} = 2$