Question

Cube root of (x+1)(y+2)(z+3)=7
Find xyz.

Answer 1

Step-by-step explanation:

• Given  
• Cube root of (x+1)(y+2)(z+3)=7 Find xyz.
• Given cube root of (x + 1)(y + 2)(z + 3) = 7
• [(x + 1)(y + 2) (z + 3) ]^1/3 = 7
• Cubing both sides we get
•  (x + 1) (y + 2) (z + 3) = 7^3
•  (x + 1) (y + 2) (z + 3) = 7 x 7 x 7 (since 7 is a prime factor)
•    x + 1 = 7
•    x = 7 – 1
•      x = 6
•     y + 2 = 7
•     y = 5
•    z + 3 = 7
•     z = 4
• Now xyz = 6 x 5 x 4  
•              = 120
• This is one solution.

Answer 2

Multiple Solutions for xyz  if ∛(x+1)(y+2)(z+3)=7 , Simplest solution is xyz = 120

Step-by-step explanation:

∛(x + 1)(y + 2)(z + 3)  = 7

Cubing both sides

=> (x + 1)(y + 2)(z + 3)  = 343

343 = 7 * 7 * 7

343 = 7 * 49 = 1 * 343   = 7 * (-7)(-7)  and so so on

Hence  (x + 1) , (y + 2) , (z + 3) will be multiple of ±7 & ±1

There can be multiple solution to this

1st simplest solution :

(x + 1) = (y + 2) = (z + 3) = 7

=> x = 6 , y = 5 , z = 4

xyz = 120

Another solution

(x + 1) = 7   (y + 2) = (z + 3) = -7  

=> x = 6  , y = -9 , z = -10

=> xyz = 540

similarly  we can have two terms - 7  & one term 7

and have different solution

Another solution

x + 1 = 1    & y+2 = 49 , z + 3 = 7  

x + 1 = 1     & y + 2 = 343  z + 3 = 1

x + 1 = 1    & y+2 = 7 , z + 3 = 49

x + 1 = 1     & y + 2 = 1  z + 3 = 343

Also y + 2 & z + 3 can be -ve signs

here x = 0

Hence xyz = 0

this way we can have a lot of solutions

Simplest one is x = 6 , y = 5 , z = 4

xyz = 120