Let the polynomial be
f(x)=ax3+bx2+cx+d
⇒f′(x)=3ax2+2bx+c
⇒f′′(x)=6ax+2b
f′′(1)=0⇒6a+2b=0⇒b=−3a
f′(−1)=0⇒3a−2b+c=0
⇒c=−9a
f(−1)=10⇒−a+b−c+d=10
⇒−a−3a+9a+d=10
d=−5a+10
f(1)=−6⇒a+b+c+d=−6
⇒a−3a−9a−5a+10=−6
⇒a=1
∴f′(x)=3x2−6x−9=3(x2−2x−3)
For f′(x)=0⇒x2−2x−3=0⇒x=3,−1
Minima exists at x=3