Question

if 21168=x^4×y^3×z^2, then. find (x+y+z)^(y+z/x+y), where x,y, and z are positive integers
PLEASE ANSWER ASAP​

Answer 1

Answer:

2√3

Step-by-step explanation:

prime factorization of 21168 = 2×2×2×2×3×3×3×7×7

= 2^4 × 3^3 × 7^2

But given that 21168 = x^4 + y^3 + z^2

on comparing , we have

x = 2 , y= 3 ,z = 7

We have to find,

( x + y + z )^(y + z/x+y)

= (2+3+7)^(3+7/2+3)

= 12^(10/5)

= 12^(1/2)

= √12

= 2√3

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