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