Given L(x)=x³+2ax+b-a find "a" and "b" if L(-2)=6 and L(1)=3

Answers:$$ a=-2 $$ $$ b=4 $$ Using the given function, substitute the given to form two equations with variables a and b. First equation:$$ 6=\left(-2\right)^3+2a\left(-2\right)+b-a $$ $$ 6=-8-4a+b-a $$ $$ 6+8=-5a+b $$ $$ -5a+b=14 $$ Second equation:$$ 3=\left(1\right)^3+2a\left(1\right)+b-a $$ $$ 3=1+2a+b-a $$ $$ 3-1=a+b $$ $$ a+b=2 $$ Now, using the two equations, solve for one by elimination.$$ \left(a+b=2\right)\left(-1\right) $$ $$ -a-b=-2 $$ Add this from the other equation to eliminate b. $$ \left(-5a+b=14\right)+(-a-b=-2) $$ Resulting to:$$ -6a=12 $$ Dividing by -6, $$ a=-2 $$ To solve for b, substitute to any of the two equation. This time I choose$$ a+b=2 $$ $$ \left(-2\right)+b=2 $$ $$ b=2+2 $$ $$ b=4 $$