Question

If x, y, z € R and 121x2 + 4y2 + 9z2 - 22x + 4y + 6z + 3 =0 then value of
xyz is equal​

Answer 1

Answer:

if x,y,z E r and 121x²+4y²+9z²-22x+4y+6z+3=0

[(11x)^2-22x+1] + [(2y)^2+4y+1]+[(3z)^2+6z+1]=0

(11x-1)^2+(2y+1)^2+(3z+1)^2

We know that squares of real no. are always positive therefore sum of their square can't be 0 unless and until they are zero

x=1/11

y=-1/2

z=-1/3

x^-1-y^-1-z^-1=11+2+3=16

Step-by-step explanation:

Answer 2

Step-by-step explanation:

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