设x1,x2是方程3x2+2x-6=0的两个根,求（1）x1的三次方+x2的三次方的值 (2)3x1的平方乘（6-2x2

（1）由韦达定理,x1+x2 = -2/3 ,x1x2 = -2
于是,
x1^3+x2^3 = (x1+x2)(x1²-x1x2+x2²) = -2/3[(x1+x2)²-3x1x2]
= -116/27
（2）3x2²+2x2-6=0,于是6-2x2 = 3 x2²
∴3x1²(6-2x2) = 9(x1x2)² = 36