Find two solutions to the equation (x^3 − 64)(x^5 − 1) = 0.

the two roots are x = 1 and x = 4

Step-by-step explanation:

Data provided in the question:

(x³ − 64) (x⁵ − 1) = 0.

Now,

for the above relation to be true the  following condition must be followed:

Either  (x³ − 64) = 0 ............(1)

or

(x⁵ − 1) = 0 ..........(2)

Therefore,

considering the first equation, we have

(x³ − 64) = 0

adding 64 both sides, we get

x³ − 64 + 64 = 0 + 64

or

x³ = 64

taking the cube root both the sides, we have

∛x³ = ∛64

or

x = ∛(4 × 4 × 4)

or

x = 4

similarly considering the equation (2) , we have

(x⁵ − 1) = 0

adding the number 1 both the sides, we get

x⁵ − 1 + 1 = 0 + 1

or

x⁵ = 1

taking the fifth root both the sides, we get

also,

1 can be written as 1⁵

therefore,

or

x = 1

Hence,

the two roots are x = 1 and x = 4

c. 16 units

step-by-step explanation:

mark brainliest and have a great day!

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.