IF f(x)=x^3-2x^3+3x^3-ax+b is divided by (x-1) and (x+1) , it leaves the remainder 5 and 19 respectively. find the values of a
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f(x)=x^3-2x^3+3x^3-ax+b
x+1=0 ,x=-1
f(-1)=-1^3-(2)-1^3+3(-1)^3-a(-1)+b
f(-1)=-1^3-2(-1)^3+3(-1)^3-a(-1)+b
f(-1)=-1+2-3+a+b
f(-1)=19
19=-2+a+b
a+b=21
if x-1=0 then x=1
f(x)=x^3-2x^3+3x^3-ax+b
f(1)=1^3-2(1)^3+3(1)^3-a(1)+b
f(1)=1-2+3-a+b
f(1)=5 then
5=2-a+b
b-a=3
a+b=21
2b=24
b=12
a=9
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