`a)Ta` `có:` `0=(a+b+c)^2`
`=>a^2+b^2+c^2=-2ab-2bc-2ca`
`\text{ Bình phương 2 vế, ta được:}`
`(a^2+b^2+c^2)^2=4(a^2b^2+b^2c^2+c^2a^2)+8(ab^2c+bc^2a+ca^b)`
`=4(a^2b^2+b^2c^2+c^2a^2`
`\text{ Mặt khác}`
`(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)`
`=>(a^2+b^2+c^2)=2(a^4+b^4+c^4)`
`b)Từ:` `a+b+c+d=0`
`=>a+b=-(c+d)`
`=>(a+b)^3=-(c+d)^3`
`=>a^3+b^3+3ab(a+b)=-c^3-d^3-3cd(c+d)`
`=>a^3+b^3+c^3+d^3=-3ab(a+b)-4cd(c+d)`
`=>a^3+b^3+c^3+d^3=3ab(c+d)-4cd(c+d)`
`=>a^3+b^3+c^3+d^3=3(c+d)(ab-cd)`