Question

If $f(x) = 64{x^3} + \dfrac{1}{{{x^3}}}$ and $\alpha ,\beta $ are the roots of $4x + \dfrac{1}{x} = 2$, then
A) $f(\alpha ) = - 64$
B) $f(\beta ) = - 8$
C) $f(\beta ) = - 16$
D) $f(\alpha ) = - 24$

Answer 1

- Hint: We will simplify f(x) by making use of formula ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$. Then we will substitute $\alpha $ and $\beta $ in the equation $4x + \dfrac{1}{x} = 2$ instead of x one by one. Then we will find the value of $f(\alpha )$ and $f(\beta )$. Then we will examine the values of $f(\alpha )$ and $f(\beta )$ , and then accordingly we will select the correct answer from the given options.

Complete step-by-step solution -
We will simplify $f(x) = 64{x^3} + \dfrac{1}{{{x^3}}}$ by converting 64 into cube of 4
$ f(x) = {(4x)^3} + \dfrac{1}{{{x^3}}}$ …… (1)
Now we will simplify further by comparing ${a^3} + {b^3} = {(a + b)^3} - 3ab(a + b)$ with equation 1, we get
$f(x) = {\left( {4x + \dfrac{1}{x}} \right)^3} - 3(4x)\left( {\dfrac{1}{x}} \right)\left[ {4x + \dfrac{1}{x}} \right]$
Now we will eliminate x and find the product of 3 & 4
$ f(x) = {\left( {4x + \dfrac{1}{x}} \right)^3} - 12\left( {4x + \dfrac{1}{x}} \right)$ …… (2)
Now we know that $\alpha ,\beta $ are the roots of $4x + \dfrac{1}{x} = 2$ , thus we can conclude that
$4\alpha + \dfrac{1}{\alpha } = 2$ and …… (3)
\[\;4\beta + \dfrac{1}{\beta } = 2\;\] …… (4)
Now we will find the value of $f(\alpha )$ by substituting x = $\alpha $ in equation 2
$ f(\alpha ) = {\left( {4\alpha + \dfrac{1}{\alpha }} \right)^3} - 12\left( {4\alpha + \dfrac{1}{\alpha }} \right)$
Now we substitute the value of $4\alpha + \dfrac{1}{\alpha } = 2$ in the above equation to simply it further
$ f(\alpha ) = {2^3} - 12(2)$
Now simplify further by finding the cube of 2
 $ f(\alpha ) = 8 - 12(2)$
Now simplify further by finding the product of 12 & 2
$ f(\alpha ) = 8 - 24$
Now simply further by finding the difference of 8 & 24 thereby finding the value of $f(\alpha )$
$ f(\alpha ) = - 16$ …… (5)
Now we will find the value of $f(\beta )$ by substituting x = $\beta $ in equation 2
$ f(\beta ) = {\left( {4\beta + \dfrac{1}{\beta }} \right)^3} - 12\left( {4\beta + \dfrac{1}{\beta }} \right)$
Now we substitute the value of \[\;4\beta + \dfrac{1}{\beta } = 2\;\] in the above equation to simply it further
$ f(\beta ) = {2^3} - 12(2)$
Now simplify further by finding the cube of 2
 $ f(\beta ) = 8 - 12(2)$
Now simplify further by finding the product of 12 & 2
$ f(\beta ) = 8 - 24$
Now simply further by finding the difference of 8 & 24 thereby finding the value of $f(\beta )$
$ f(\beta ) = - 16$ …… (6)

Thus, after comparing equation 5 & 6 and options given in the question, we can conclude option C) is the right answer.

Note:
Aliter method:
We will cube the equation $4x + \dfrac{1}{x} = 2$ from both side
$ {\left( {4x + \dfrac{1}{x}} \right)^3} = {2^3}$
Now we will apply the formula of ${(a + b)^3} = {a^3} + 3{a^2}b + 3a{b^2} + {b^3}$ in the left- hand side of the above equation
$ {(4x)^3} + 3{(4x)^2}\left( {\dfrac{1}{x}} \right) + 3(4x){\left( {\dfrac{1}{x}} \right)^2} + {\left( {\dfrac{1}{x}} \right)^3} = {2^3}$
Now we will solve the exponents in the above equation
$ 64{x^3} + 3\left( {16{x^2}} \right)\left( {\dfrac{1}{x}} \right) + 3(4x)\left( {\dfrac{1}{{{x^2}}}} \right) + \left( {\dfrac{1}{{{x^3}}}} \right) = 8$
Now we will find the product of 3 and $16{x^2}$,and 3 and $4x$
$ 64{x^3} + 48{x^2}\left( {\dfrac{1}{x}} \right) + 12x\left( {\dfrac{1}{{{x^2}}}} \right) + \left( {\dfrac{1}{{{x^3}}}} \right) = 8$
Now we will reduce ${x^2}$ and $x$
$ 64{x^3} + 48x \times 1 + 12\left( {\dfrac{1}{x}} \right) + \left( {\dfrac{1}{{{x^3}}}} \right) = 8$
$ 64{x^3} + 48x + \left( {\dfrac{{12}}{x}} \right) + \left( {\dfrac{1}{{{x^3}}}} \right) = 8$
Now we will arrange the terms on the left-hand side in such a way that cube terms come together
$ 64{x^3} + \dfrac{1}{{{x^3}}} + 48x + \dfrac{{12}}{x} = 8$
Now we will take 12 common from the expression $48x + \dfrac{{12}}{x}$
$ 64{x^3} + \dfrac{1}{{{x^3}}} + 12\left( {4x + \dfrac{1}{x}} \right) = 8$
Now we know that $f(x) = 64{x^3} + \dfrac{1}{{{x^3}}}$ and $\left( {4x + \dfrac{1}{x}} \right) = 2$ so our above equation becomes
$ f(x) + 12(2) = 8$
Now we will find the product of 12 and 2
\[ f(x) + 24 = 8\]
Now we will transpose 24 on the right-hand side
\[ f(x) = 8 - 24\]
Now we will find the difference of 8 and 24
\[ f(x) = - 16\]
Now it is given that $\alpha $ and $\beta $ are the roots of the equation $\left( {4x + \dfrac{1}{x}} \right) = 2$ and we are solving the same equation. So, we can say that \[f(\alpha ) = f(\beta ) = - 16\]

Hence, the option C) is correct.