Question

if x^2,y^2,z^2 = 3^2×5^2×7^2 then find (x+y+z)​

Answer 1

Answer:

x^2 +y^2+z^2=2(x-y-z)-3

Or, x^2+y^2+z^2=2x-2y-2z-3

Or, x^2+y^2+z^2–2x+2y+2z+3=0

Or,(x^2–2x+1)+(y^2+2y+1)+(z^2+2z+1)=0

Or, (x-1)^2+(y+1)^2+(z+1)^2=0

Now the sum of the squares is 0

only when they individually becomes 0.

So, (x-1)^2=0

Or, x-1=0

Or, x=1

(y+1)^2=0

Or, y+1=0

Or, y=-1

And (z+1)^2=0

Or, z+1=0

Or, z=-1

Now 2x-3y+4z

=2×1–3×(-1)+4×(-1)

=2+3–4

=5–4

=1

Answer 2

Step-by-step explanation:

refer to the attachment

hope it helps :)