Question

If ${x^{ - 2}} = 64$ then what will be the value of $\dfrac{1}{{{x^3}}} + {x^0}$

Answer 1

Hint: We need to find the value of $x$ using the normal square root method. Then the value of $x$ is a substituted value. Then we just need to put the value of x in the given expression to get our required answer.

Complete step-by-step answer:
Given,
The value of x is given in an equation.
The equation given is ${x^{ - 2}} = 64$.
By finding the value $x$from the above equation, then we can substitute the value of $x$in the equation $\dfrac{1}{{{x^3}}} + {x^0}$
We need to find the value of $x$from the equation,
${x^{ - 2}} = 64$
To eliminate the negative value of power we can multiply both sides the $^{ - 1}$In the power,
${x^{ - 2 \times - 1}} = {64^{ - 1}}$
We can multiply the above equation,
${x^2} = {64^{ - 1}}$
If the power is negative then, it can be removed by bringing the value in the numerator to the denominator, so that the negative value of power can be eliminated,
${x^2} = \dfrac{1}{{64}}$
Now we can multiply power $\dfrac{1}{2}$on both sides of the equation,
${x^{2 \times \dfrac{1}{2}}} = {\left( {\dfrac{1}{{64}}} \right)^{\dfrac{1}{2}}}$
Canceling the power terms
$x = \dfrac{1}{{{{64}^{\dfrac{1}{2}}}}}$
$64$can be written as $8 \times 8$
$x = \dfrac{1}{{{{(8 \times 8)}^{\dfrac{1}{2}}}}}$
As the base are equal, then we can add the value in power,
$x = \dfrac{1}{{{{({8^{1 + 1}})}^{\dfrac{1}{2}}}}}$
Add the values in power
$x = \dfrac{1}{{{{({8^2})}^{\dfrac{1}{2}}}}}$
By canceling the values in power since they are equal.
$x = \dfrac{1}{8}$
Finally, we found the value of $x$, such that $x = \dfrac{1}{8}$
Now we need to find the value of $\dfrac{1}{{{x^3}}} + {x^0}$
We need to substitute the value of $x$in the above equation,
$\dfrac{1}{{{x^3}}} + {x^0} = \dfrac{1}{{{{\left( {\dfrac{1}{8}} \right)}^3}}} + {\left( {\dfrac{1}{8}} \right)^0}$
Any value to the power $0$is equal to $1$, such that ${a^0} = 1$
$\dfrac{1}{{{x^3}}} + {x^0} = \dfrac{1}{{{{\left( {\dfrac{1}{8}} \right)}^3}}} + 1$
By multiplying the values in power to the base,
$\dfrac{1}{{{x^3}}} + {x^0} = \dfrac{1}{{\left( {\dfrac{{{1^3}}}{{{8^3}}}} \right)}} + 1$
The value in the denominator can be brought to the numerator,
$\dfrac{1}{{{x^3}}} + {x^0} = \dfrac{{{8^3}}}{{{1^3}}} + 1$
Any value in the numerator with the denominator $1$can be written only numerator,
$\dfrac{1}{{{x^3}}} + {x^0} = {8^3} + 1$
The power is being multiplied in the above equation,
$\dfrac{1}{{{x^3}}} + {x^0} = 512 + 1$
The addition is done on the above equation,
$\dfrac{1}{{{x^3}}} + {x^0} = 513$
We found the value of $\dfrac{1}{{{x^3}}} + {x^0} = 513$.

Note: The value must be found correctly, then only by substituting the value we can find the correct answer.
Any value to the power $0$is equal to $1$ is not $a$ such that ${a^0} = 1$not ${a^0} \ne a$. This must be taken into consideration. The main mistake done will be negative and should be taken as reciprocal.