If x2+y2+z2=2(x-y-z)-3 find the value of 2x-3y+4z
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Given x+y+z=0
Then the value of (x^2y^2+y^2z^2+z^2x^2) ÷ (x^4+y^4+z^4)
Put x=1, y=-1 and z=0 in the given
1-1+0=0
Now ,
(x^2y^2+y^2z^2+z^2x^2) ÷ (x^4+y^4+z^4)
(1^2(-1)^2+(-1)^2(0)^2+(0)^2(1)^2) ÷ ((1)^4+(-1)^4+0^4)
(1+0+0)/(1+1+0)
1/2
Then the value of (x^2y^2+y^2z^2+z^2x^2) ÷ (x^4+y^4+z^4)
Put x=1, y=-1 and z=0 in the given
1-1+0=0
Now ,
(x^2y^2+y^2z^2+z^2x^2) ÷ (x^4+y^4+z^4)
(1^2(-1)^2+(-1)^2(0)^2+(0)^2(1)^2) ÷ ((1)^4+(-1)^4+0^4)
(1+0+0)/(1+1+0)
1/2
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