# If x 2 +y 2 +z 2 = 121, xyz= 108 and (1/x +1/y + 1/z)=7/9, where x>0,

| If x

^{2}+y^{2}+z^{2}= 121, xyz= 108 and (1/x +1/y + 1/z)=7/9, where x>0, y> 0 and z>0; what is the value of x+ y+ z?A. ±17

B. 17

C. 18

D. ±18

### Right Answer is: B

#### SOLUTION

we know that ,

it is given that 1/x +1/y +1/z =7/9

(xyz=108 given)

now putting the values in we get

= 121+2×84=121+168=289

x+ y+ z= ±17 but negative value of 17 can not be considered here because it is given that x, y and z are greater than zero.

hence x+ y+ z= 17

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