一道数学题!谢谢

当x=-3时,ax^3+bx^3+cx-5=7,求当x=3时,ax^5+bx^3+cx-5的值

ax^3+bx^3+cx-5=7应该是ax^5+bx^3+cx-5=7吧

ax^5+bx^3+cx-5=7

即a(-3)^5+b(-3)^3+c(-3)-5=7

即-a3^5-b3^3-3c-5=7

即a3^5+b3^3+3c=-12

当x=3时,ax^5+bx^3+cx-5=a3^5+b3^3+3c-5
=-12-5
=-17

a(-3)^3+b(-3))^3+c(-3)-5=7
a(-3)^3+b(-3))^3+c(-3)=12
a(3)^3+b(3))^3+c(3)=-12
a(3)^3+b(3))^3+c(3)-5=-12-5=-17

看不懂…

依题意有ax^3+bx^3+cx=12
X=-3时成立
X=3时...ax^3+bx^3+cx=-12
ax^5+bx^3+cx-5=-12-5=-17

难