A, b, c, d are the roots of 3x⁴-6x²+2=0. what is the value of a⁴+b⁴+c⁴+d⁴?

Hello,

P(x)=x^4-6x²+2=(x-a)(x-b)(x-c)(x-d)
=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2-(abc+abd+acd+bcd)x+abcd

==>
ab+ac+ad+bc+bd+cd=-6
abc+abd+acd+bcd=0
abcd=2
a+b+c+d=0 ==>(a+b+c+d)²=0=a²+b²+c²+d²+2(ab+ac+ac+bc+bd+cd)
==>a²+b²+c²+d²=0-2*(-6)=12

if a is a root P(a)=0==>a^4-6a²+2=0
if b is a root P(b)=0==>b^4-6b²+2=0
if c is a root P(a)=0==>c^4-6c²+2=0
if d is a root P(a)=0==>d^4-6d²+2=0

==>a^4+b^4+c^4+d^4-6(a²+b²+c²+d²)+4*2=0
==>a^4+b^4+c^4+d^4=-8+6*12=64