Find the value of z such that 2(z - 1)^3 + 6(1 - z)^3 = 32?

Question

Find the value of z such that  2(z - 1)^3 + 6(1 - z)^3 = 32 ?

A. -2
B. -1
C. 0
D. 1
E. 2

rajatchopra1994 · Answer

Explanation:

2*(z−1)^3 + 6*(1−z)^3 = 32

We can solve this by putting values of z, as it will be less time-consuming.

So when z =-1

2(-2)^3 + 6(2)^3
2*(-8) + 6*8
-16 + 48
32

IMO-B

CrackverbalGMAT · Answer

2(z - 1)^3 + 6(1 - z)^3 = 32

Changing the sign for the second term, we get

2(z - 1)^3 - 6(z - 1)^3 = 32

(z - 1)^3 (2 - 6) = 32

(z - 1)^3 = \frac{32}{-4} = -8

z - 1 = -2

z = -1

Option B

Arun Kumar

gmatcafe · Answer

2(z - 1)^3 + 6(1 - z)^3 = 32 
 -2(1 - z)^3 + 6(1 - z)^3 = 32 
 4(1 - z)^3 = 32 
 (1 - z)^3 = 8 
 1 - z = 2  
 z = -2  B)

winterschool · Answer

2(z - 1)^3 + 6(1 - z)^3 = 32

2(z - 1)^3 + 6 (-1)^3 * (z-1)^3 = 32

2(z - 1)^3 - 6 * (z-1)^3 = 32

- 4 (z-1)^3 = 32

(z-1)^3 = -8

(1-z)^3 = 8

1-z = 2

z = -1

Answer B

bluejeanbaby · Answer

Use the identity with the blue arrow to reduce this down -- no need to plug in answer choices! This is the most elegant way. Best solution.png

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