ANSWERS & EXPLANATIONS–

665. b.We must first rewrite the equation in a nicer form. Observe that

3 +x

• 41 
  
  
  • x
    
    
    • 21 
      = 0
    
    
    
    
  
  
  
  


• x


• 21 
  +x


• 41 
  = 0


• (x


• 41 
  )^2 + (x


• 41 
  ) + 3 = 0

Let u= x

• 41 
  . Then, solve the quadratic equation


• u^2 + u+ 3 = 0. Using the quadratic formula yields

u= = =.

Now, we must go back to the actual substitution and solve the following equations involving the original

variable x:

x

• 41 
  = x
  
  
  • 
    

    41 

    
    
  
  
  
  

(x

• 41 
  )–4= ( )–4 (x
  
  
  • 41 
    )–4= ( )–4
  
  
  
  

x= ( )^4 = x= ( )^4 =

So, the two solutions to the original equation are x= ,.

666. d.Let u = x^3 + 5. Then, the equation (x^3 + 5)^2 – 5(x^3 + 5) + 6 = 0 can be written equivalently as u^2 – 5u+
     6 = 0. This factors as (u– 3)(u– 2) = 0, so we conclude that u= 3, 2. Next, we must solve the following
     equations obtained by going back to the actual substitution:

x^3 + 5 = 3 x^3 + 5 = 2

x^3 = –2 x^3 = –3

(x^3 )

(^13) 
= (–2)
(^13) 
(x^3 )
(^13) 
= (–3)
(^13) 
x= ^3 –2 x= ^3 –3
So, the two solutions to the original equation are x= ^3 –2,^3 –3.

667. c.Observe that 4x^6 + 1 = 5x^3 , or equivalently 4x^6 – 5x^3 + 1 = 0, can be written as 4(x^3 )^2 – 5(x^3 ) + 1 = 0.
     Let u = x^3. Rewriting the original equation yields the equation 4|u|^2 – 5|u| + 1 = 0, which is quadratic.
     Factoring yields the equivalent equation (4u – 1)(u – 1) = 0. Solving this equation for uyields the solu-
     tions u = ^14 or u = 1. Solving the original equation requires that we go back to the substitution and write
     u in terms of the original variable x:

u= ^14 is the same as x^3 = ^14 , so that x=    3 

u= 1 is the same as x^3 = 1, so that x= 1

The solutions of the original equation arex= 3 ^14 ,1.

^1

4

^16

(1 –  13 )^4

^16

(1 +  13 )^4

^16

(1 –  13 )^4

^2

1 –  13 

^16

(1 +  13 )^4

^2

1 +  13 

1 – ^13 

2

1 + ^13 

2

1 – ^13 

2

1 + ^13 

2

1 ±^13 

2

–1 ±^13 

–2

–1 ±1 – 4(–1)(3)

2(–1)