Answer:
So in standard form equation will be [tex]x^2+2x-1[/tex]
Step-by-step explanation:
We have given expression [tex]\frac{x^2-5x+6}{x-3}+\frac{x^3-1}{x-1}[/tex]
Let first we solve first part of the expression
So [tex]\frac{x^2-5x+6}{x-3}=\frac{x^2-3x-2x+6}{x-3}=\frac{(x-3)(x-2)}{x-3}=x-2[/tex]
Now second part [tex]\frac{x^3-1}{x-1}[/tex]
We know the algebraic identity [tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
So by using this identity
[tex]\frac{x^3-1}{x-1}=\frac{(x-1)(x^2+x+1)}{x-1}=x^2+x+1[/tex]
Now adding first and second part
[tex]x-2+x^2+x+1=x^2+2x-1[/tex]
So in standard form equation will be [tex]x^2+2x-1[/tex]