Question

If x-1/x=5 find x^3+1/x^3=?

Answer 1

To solve this problem, we need to solve for x in the first equation, then substitute in this value for x into the second equation.

Let's begin by starting to solve for x in the first equation, x-1/x = 5. To begin, we should multiply both sides of the equation by x to get rid of the denominator on the left side of the equation.

x-1/x = 5

x-1 = 5x

Then, we should subtract x from both sides of the equation so that all of the variables are on the right side of the equation.

-1 = 4x

Finally, we can divide both sides of the equation by 4, to get the variable x alone on the right side.

x = -1/4

Now, we should substitute in this value of x into the second equation.

x^3 + 1 / x^3

(-1/4)^3 + 1 / (-1/4)^3

To simplify, we can begin by simplifying the exponents.

-1/64 + 1 / -1/64

Next, we should change 1 into 64/64 so that we can simplify the numerator of the fraction.

-1/64 + 64/64 / -1/64

To simplify, we need to add the two fractions in the numerator.

63/64 / -1/64

Because both of the "individual" fractions of the numerator and denominator have the common denominator of 64, we can get rid of the both of the denominators, as follows:

63/-1

Finally, we can perform this simple division.

-63

Therefore, your answer is -63.

Hope this helps!

Answer 2

$\dfrac{x-1}{x}=5\ \ \ \ /x\neq0\\\\\dfrac{x-1}{x}=\dfrac{5}{1}\ \ \ \ |\text{cross multiply}\\\\5x=x-1\ \ \ \ |-x\\\\5x=1\ \ \ \ |:5\\\\x=0.2\\\\\dfrac{x^3+1}{x^3}=\dfrac{x^3}{x^3}+\dfrac{1}{x^3}=1+\dfrac{1}{x^3}\to1+\dfrac{1}{0.2^3}=1+\dfrac{1}{0.008}\\\\=1+\dfrac{1}{\frac{8}{1,000}}=1+\dfrac{1,000}{8}=1+125=126\\\\\text{Answer:}\ \dfrac{x^3+1}{x^3}=126$